Proving that if x^4 + 5x + 1 < 27 then x < 2

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I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q



P : x^4 + 5x + 1 < 27

Q : x < 2


I wanted to try and prove this by contrapositive , so this state would become



If X>=2 then x^4 + 5x + 1 >= 27


Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.



Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.










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  • No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
    – Heisenberg
    1 hour ago










  • If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
    – copper.hat
    1 hour ago










  • You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
    – Hagen von Eitzen
    1 hour ago














up vote
2
down vote

favorite












I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q



P : x^4 + 5x + 1 < 27

Q : x < 2


I wanted to try and prove this by contrapositive , so this state would become



If X>=2 then x^4 + 5x + 1 >= 27


Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.



Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.










share|cite|improve this question





















  • No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
    – Heisenberg
    1 hour ago










  • If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
    – copper.hat
    1 hour ago










  • You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
    – Hagen von Eitzen
    1 hour ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q



P : x^4 + 5x + 1 < 27

Q : x < 2


I wanted to try and prove this by contrapositive , so this state would become



If X>=2 then x^4 + 5x + 1 >= 27


Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.



Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.










share|cite|improve this question













I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form P -> Q



P : x^4 + 5x + 1 < 27

Q : x < 2


I wanted to try and prove this by contrapositive , so this state would become



If X>=2 then x^4 + 5x + 1 >= 27


Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me 2^4 + 5*2 + 1 = 27 and since 27>=27 this statement is True.



Am I allowed to prove it like this? Is there a different way to prove a question like this? Any help would be appreciated thank you.







proof-verification proof-writing proof-explanation






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asked 1 hour ago









Rufyi

355




355











  • No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
    – Heisenberg
    1 hour ago










  • If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
    – copper.hat
    1 hour ago










  • You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
    – Hagen von Eitzen
    1 hour ago
















  • No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
    – Heisenberg
    1 hour ago










  • If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
    – copper.hat
    1 hour ago










  • You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
    – Hagen von Eitzen
    1 hour ago















No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago




No, you have to prove for a general $xgeq 2$. It might work for a specific value but not in general. That is the problem with your approach
– Heisenberg
1 hour ago












If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago




If $x ge 2$ then $x^4 + 5x +1 ge 16+10+1 = 27$.
– copper.hat
1 hour ago












You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago




You can use that for $cge0$, $age b$ implies $acge bc$. So for $xge 2$ (and thereby $x>0$) we have $x^2ge 2xge 4$, $x^3ge 2x^2ge 8$, $x^4ge 2x^3ge 16$, and hence by addimg the inequality $x^4+4x+1ge 16+10+1=27$ indeed
– Hagen von Eitzen
1 hour ago










4 Answers
4






active

oldest

votes

















up vote
3
down vote



accepted










Suppose $xgeq2$. Then $x^4geq16implies x^4
+5xgeq16+5xgeq 16+...$
. Got the hint?






share|cite|improve this answer




















  • Are you trying to make the left side look like what we need to prove?
    – Rufyi
    1 hour ago










  • If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
    – Rufyi
    1 hour ago










  • Yes exactly!...
    – Heisenberg
    1 hour ago










  • Can I do what I said in the second comment?
    – Rufyi
    1 hour ago










  • Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
    – Heisenberg
    1 hour ago


















up vote
1
down vote













A marginally different take:



Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.



Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.



In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
$x < 2$.






share|cite|improve this answer



























    up vote
    0
    down vote













    Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that



    $$xgeq 2implies x^4+5x+1leq 27,$$



    which is false for example at $x=0$, you can't just give the example of $x=2$.



    One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.






    share|cite|improve this answer



























      up vote
      0
      down vote













      If $xge 2$ is it true that $x^4 ge 16$?



      If $x ge 2$ is it true that $5x ge 10$?



      So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?



      ...



      It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.




      This may be tedious over kill but:



      If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$



      Now because $d ge 0$ then $d^k ge 0$.



      So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$



      And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$



      So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.



      That should do it.....






      share|cite|improve this answer






















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        Suppose $xgeq2$. Then $x^4geq16implies x^4
        +5xgeq16+5xgeq 16+...$
        . Got the hint?






        share|cite|improve this answer




















        • Are you trying to make the left side look like what we need to prove?
          – Rufyi
          1 hour ago










        • If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
          – Rufyi
          1 hour ago










        • Yes exactly!...
          – Heisenberg
          1 hour ago










        • Can I do what I said in the second comment?
          – Rufyi
          1 hour ago










        • Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
          – Heisenberg
          1 hour ago















        up vote
        3
        down vote



        accepted










        Suppose $xgeq2$. Then $x^4geq16implies x^4
        +5xgeq16+5xgeq 16+...$
        . Got the hint?






        share|cite|improve this answer




















        • Are you trying to make the left side look like what we need to prove?
          – Rufyi
          1 hour ago










        • If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
          – Rufyi
          1 hour ago










        • Yes exactly!...
          – Heisenberg
          1 hour ago










        • Can I do what I said in the second comment?
          – Rufyi
          1 hour ago










        • Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
          – Heisenberg
          1 hour ago













        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Suppose $xgeq2$. Then $x^4geq16implies x^4
        +5xgeq16+5xgeq 16+...$
        . Got the hint?






        share|cite|improve this answer












        Suppose $xgeq2$. Then $x^4geq16implies x^4
        +5xgeq16+5xgeq 16+...$
        . Got the hint?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Heisenberg

        1,3911639




        1,3911639











        • Are you trying to make the left side look like what we need to prove?
          – Rufyi
          1 hour ago










        • If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
          – Rufyi
          1 hour ago










        • Yes exactly!...
          – Heisenberg
          1 hour ago










        • Can I do what I said in the second comment?
          – Rufyi
          1 hour ago










        • Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
          – Heisenberg
          1 hour ago

















        • Are you trying to make the left side look like what we need to prove?
          – Rufyi
          1 hour ago










        • If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
          – Rufyi
          1 hour ago










        • Yes exactly!...
          – Heisenberg
          1 hour ago










        • Can I do what I said in the second comment?
          – Rufyi
          1 hour ago










        • Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
          – Heisenberg
          1 hour ago
















        Are you trying to make the left side look like what we need to prove?
        – Rufyi
        1 hour ago




        Are you trying to make the left side look like what we need to prove?
        – Rufyi
        1 hour ago












        If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
        – Rufyi
        1 hour ago




        If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side?
        – Rufyi
        1 hour ago












        Yes exactly!...
        – Heisenberg
        1 hour ago




        Yes exactly!...
        – Heisenberg
        1 hour ago












        Can I do what I said in the second comment?
        – Rufyi
        1 hour ago




        Can I do what I said in the second comment?
        – Rufyi
        1 hour ago












        Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
        – Heisenberg
        1 hour ago





        Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1geq 27$
        – Heisenberg
        1 hour ago











        up vote
        1
        down vote













        A marginally different take:



        Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.



        Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.



        In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
        we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
        $x < 2$.






        share|cite|improve this answer
























          up vote
          1
          down vote













          A marginally different take:



          Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.



          Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.



          In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
          we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
          $x < 2$.






          share|cite|improve this answer






















            up vote
            1
            down vote










            up vote
            1
            down vote









            A marginally different take:



            Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.



            Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.



            In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
            we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
            $x < 2$.






            share|cite|improve this answer












            A marginally different take:



            Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.



            Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.



            In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x ge 0$,
            we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have
            $x < 2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            copper.hat

            124k558156




            124k558156




















                up vote
                0
                down vote













                Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that



                $$xgeq 2implies x^4+5x+1leq 27,$$



                which is false for example at $x=0$, you can't just give the example of $x=2$.



                One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.






                share|cite|improve this answer
























                  up vote
                  0
                  down vote













                  Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that



                  $$xgeq 2implies x^4+5x+1leq 27,$$



                  which is false for example at $x=0$, you can't just give the example of $x=2$.



                  One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.






                  share|cite|improve this answer






















                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that



                    $$xgeq 2implies x^4+5x+1leq 27,$$



                    which is false for example at $x=0$, you can't just give the example of $x=2$.



                    One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.






                    share|cite|improve this answer












                    Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that



                    $$xgeq 2implies x^4+5x+1leq 27,$$



                    which is false for example at $x=0$, you can't just give the example of $x=2$.



                    One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Carl Schildkraut

                    9,82911238




                    9,82911238




















                        up vote
                        0
                        down vote













                        If $xge 2$ is it true that $x^4 ge 16$?



                        If $x ge 2$ is it true that $5x ge 10$?



                        So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?



                        ...



                        It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.




                        This may be tedious over kill but:



                        If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$



                        Now because $d ge 0$ then $d^k ge 0$.



                        So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$



                        And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$



                        So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.



                        That should do it.....






                        share|cite|improve this answer


























                          up vote
                          0
                          down vote













                          If $xge 2$ is it true that $x^4 ge 16$?



                          If $x ge 2$ is it true that $5x ge 10$?



                          So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?



                          ...



                          It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.




                          This may be tedious over kill but:



                          If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$



                          Now because $d ge 0$ then $d^k ge 0$.



                          So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$



                          And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$



                          So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.



                          That should do it.....






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            If $xge 2$ is it true that $x^4 ge 16$?



                            If $x ge 2$ is it true that $5x ge 10$?



                            So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?



                            ...



                            It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.




                            This may be tedious over kill but:



                            If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$



                            Now because $d ge 0$ then $d^k ge 0$.



                            So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$



                            And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$



                            So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.



                            That should do it.....






                            share|cite|improve this answer














                            If $xge 2$ is it true that $x^4 ge 16$?



                            If $x ge 2$ is it true that $5x ge 10$?



                            So if $x ge 2$ is it true that $x^4 + 5x + 1 ge 16 + 10 + 1 = 27$?



                            ...



                            It could get tedious to go to axioms and prove that if $xge 2 > 0$ then $x^4 ge 2x^3 ge 4x^2 ge 8x ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $xge 2 implies 5x ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c iff a > b$ then $x^4 + 5x + 1 ge x^4 + 10 + 1 ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.




                            This may be tedious over kill but:



                            If $x ge 2$ then let $d = x - 2 ge 0$. Then $x = 2 + d$



                            Now because $d ge 0$ then $d^k ge 0$.



                            So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 ge 2^4$



                            And $5x = 5*(2+d) = 5*2 + 5*d ge 5*2$



                            So $x^4 + 5x + 1ge 2^4 + 5*2 + 1 = 27$.



                            That should do it.....







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 39 mins ago

























                            answered 51 mins ago









                            fleablood

                            63.4k22679




                            63.4k22679



























                                 

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