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Equation

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Lê Anh Duy
04/03/2019 at 15:28
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3
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32) Solve the equation (And you will know the surprising age of Tran Anh):

\(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102\)

(P/s: Please do this math problem by yourself, step-to-step, and don't try to copy from the Internet, please)

 

Equation

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    FacuFeri 06/03/2019 at 05:13

    We have : \(x^2-180x+8102=\left(x^2-180x+8100\right)+2=\left(x-90\right)^2+2\ge2\forall x\left(1\right)\)

    Applying the Bunhiacopxki inequality , we have : 

    \(\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le\left(1+1\right)\left(x-89+91-x\right)\)

    \(\Rightarrow\left(\sqrt{x-89}+\sqrt{91-x}\right)^2\le2.2=4\)

    \(\Rightarrow\sqrt{x-89}+\sqrt{91-x}\le2\left(2\right)\)

    Because \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102\)

    So ( 1 ) ; ( 2 ) ; we have : \(\sqrt{x-89}+\sqrt{91-x}=x^2-180x+8102=2\)

    Equal sign occurs \(\Leftrightarrow\sqrt{x-89}=\sqrt{91-x};x-90=0\left(3\right)\)

    We have : \(\sqrt{x-89}=\sqrt{91-x}\)  \(\Leftrightarrow x-89=91-x\Leftrightarrow x=90\left(4\right)\)

     ( 3 ) ; ( 4 ) \(\Rightarrow x=90\) is the result of the equation

    Lê Anh Duy selected this answer.
  • ...
    Huy Toàn 8A (TL) 06/03/2019 at 13:14

    Đk : \(89\le x\le91\)

    Applying the Bunhiacopxki inequality

    We have : VT \(=1.\sqrt{x-89}+1.\sqrt{91-x}\le\sqrt{\left(1+1\right)\left(x-89+91-x\right)=2}\)

    => VT \(\le2\)

    And VP = \(x^2-2.x.90+90^2+2=\left(x-90\right)^2+2\ge2\)

    => VP  \(\ge2\ge\) VT

    The sign "=" occurs when and only when: 

    \(\left\{{}\begin{matrix}\sqrt{x-89}=\sqrt{91-x}\\x-90=0\end{matrix}\right.\) => \(x=90\)(satisfy)

    The answer is 90

  • ...
    FacuFeri 05/03/2019 at 05:10

    Tran Anh ??? Le Quoc Tran Anh ??? 


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Quoc Tran Anh Le Coordinator
05/08/2018 at 03:46
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0
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The line with equation ax + by = c, where a, b and c are positive, forms a right triangle with legs on the x- and y-axes. What is the area of the triangle? Express your answer as a common fraction in terms of a, b and c.

MathcountsTrianglesEquation


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Uchiha Sasuke
29/06/2018 at 09:45
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0
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Given \(f\left(x\right)=\dfrac{2x^2+3x+3}{2x-1}\). If a and f (a) are both integers, we say that f (a) is an element of the set A. What is the sum of all elements of the set A?

Equation


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Lê Quốc Trần Anh Coordinator
11/07/2017 at 09:50
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1
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How many numbers \(x\) divided by 11 satisfy this equation:

\(999\le x\le1111\)

Equation

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    Phan Thanh Tinh Coordinator 11/07/2017 at 09:54

    The smallest and the biggest value of x are 1001 and 1111.The number of values of x is : (1111 - 1001) : 11 + 1 = 11

    Lê Quốc Trần Anh selected this answer.

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Dung Trần Thùy
19/03/2017 at 21:56
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0
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Help me, I have to finish my homework today :

Solve the equations :

\(\dfrac{x^2-2x+2}{x-1}+\dfrac{x^2-8x+20}{x-4}=\dfrac{x^2-6x+16}{x-3}+\dfrac{x^2-4x+2}{x-2}\)

I know the answer is 10 000 000 but I don't know how to solve this problem.

EquationSolve


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Min Hoang moderators
16/03/2017 at 11:23
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2
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If \(f\left(x\right)=\dfrac{2x+3}{2-3x},\) find all x for which \(f^{-1}\left(x\right)=2f\left(x^{-1}\right).\)

Equation

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    mathlove 16/03/2017 at 14:00

    With the condition \(x\ne0\), \(f\left(x^{-1}\right)=\dfrac{2x^{-1}+3}{2-3x^{-1}}=\dfrac{2+3x}{2x-3}\). The done equation become

                                                 \(\dfrac{2-3x}{2x+3}=\dfrac{2\left(2+3x\right)}{2x-3}\) .

    With the additional condition \(x\ne\pm\dfrac{3}{2}\) , the above equation equivalent to  

                            \(2\left(2+3x\right)\left(2x+3\right)=\left(2-3x\right)\left(2x-3\right)\)\(\Leftrightarrow18x^2-13x+18=0\)

    This equation has no solution.

    Selected by MathYouLike
  • ...
    FA KAKALOTS 28/01/2018 at 22:11

    With the condition x≠0, f(x−1)=2x−1+32−3x−1=2+3x2x−3

    . The done equation become

                                                 2−3x2x+3=2(2+3x)2x−3

     .

    With the additional condition x≠±32

     , the above equation equivalent to  

                            2(2+3x)(2x+3)=(2−3x)(2x−3)

    ⇔18x2−13x+18=0

    This equation has no solution.


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Min Hoang moderators
16/03/2017 at 10:34
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2
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Solve for x:

\(\dfrac{1}{x^2+18x-20}+\dfrac{1}{x^2+18x-24}=\dfrac{-12}{x^2+18x-29}\)

Equation

  • ...
    mathlove 16/03/2017 at 11:33

    Put  \(t=x^2+18x-24\Leftrightarrow x^2+18x-24-t=0\Leftrightarrow x=-9\pm\sqrt{105+t},\left(t\ge-105\right)\), the equation become    

                                                      \(\dfrac{1}{t+4}+\dfrac{1}{t}=\dfrac{-12}{t-5}\)  (1)

    With the condition  \(t\notin\left\{0;-4;5\right\}\), \(\left(1\right)\Leftrightarrow7t^2+21t-10=0\Leftrightarrow t=\dfrac{-21\pm\sqrt{721}}{14}\)(both of these solutions are not smaller  -105). So, the done equation have 4 solutions

                     \(x=-9\pm\pm\sqrt{\dfrac{1449+\sqrt{721}}{14}};x=-9\pm\sqrt{\dfrac{1449-\sqrt{721}}{14}}\)  .

    ​ ​

    Selected by MathYouLike
  • ...
    FA KAKALOTS 28/01/2018 at 22:11

    put  t=x2+18x−24⇔x2+18x−24−t=0⇔x=−9±√105+t,(t≥−105)

    , the equation become    

                                                      1t+4+1t=−12t−5

      (1)

    With the condition  t∉{0;−4;5}

    , (1)⇔7t2+21t−10=0⇔t=−21±√72114

    (both of these solutions are not smaller  -105). So, the done equation have 4 solutions

                     x=−9±±√1449+√72114;x=−9±√1449−√72114

      .


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Min Hoang moderators
16/03/2017 at 10:16
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0
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Solve the equation \(\sqrt{3+\sqrt{3-\sqrt{3+\sqrt{3-x}}}}=x\)

Equation


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John
11/03/2017 at 07:26
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2
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Let a, b and c be constants, find all solutions of the following equation:

   \(\left(x-a\right)\left(x-b\right)=\left(c-a\right)\left(c-b\right)\)

Equation

  • ...
    mathlove 11/03/2017 at 11:05

    x = c is a solution of the equation. 

    The equation is writed by  \(x^2-\left(a+b\right)x+c\left(a+b-c\right)=0\) . By the theorem Viette, the rest solution is  \(x=a+b-c\) .

    John selected this answer.
  • ...
    FA KAKALOTS 28/01/2018 at 22:11

    x = c is a solution of the equation. 

    The equation is writed by  x2−(a+b)x+c(a+b−c)=0

     . By the theorem Viette, the rest solution is  x=a+b−c .


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