Tìm x biết : a. (5x-4)^2 – 49x^2 = 0 b. x^2 +3x-10 =0

Question

Tìm x biết :
a. (5x-4)^2 – 49x^2 = 0
b. x^2 +3x-10 =0

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Edana Edana 8 tháng 2020-10-15T12:12:10+00:00 2 Answers 86 views 0

Answers ( )

  1. `a,` `(5x-4)^2-49x^2=0`

    `⇔(5x-4)^2-(7x)^2=0`

    `⇔(5x-4-7x)(5x-4+7x)=0`

    `⇔(-2x-4)(12x-4)=0`

    \(⇔\left[ \begin{array}{l}-2x-4=0\\12x-4=0\end{array} \right.⇔\left[ \begin{array}{l}x=-2\\x=\dfrac13\end{array} \right.\) 

    `b,` `x^2+3x-10=0`

    `⇔x^2-2x+5x-10=0`

    `⇔x(x-2)+5(x-2)=0`

    `⇔(x+5)(x-2)=0`

    \(⇔\left[ \begin{array}{l}x+5=0\\x-2=0\end{array} \right.⇔\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\) 

  2. Đáp án:

    a)\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{3}\end{array} \right.\) 

    b)\(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\) 

    Giải thích các bước giải:

     $a)(5x-4)^2-49x^2=0$

    $⇔(5x-4-7x).(5x-4+7x)=0$

    $⇔(-2x-4).(12x-4)=0$

    ⇔\(\left[ \begin{array}{l}-2x-4=0\\12x-4=0\end{array} \right.\) 

    ⇔\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{3}\end{array} \right.\) 

    $b)x^2+3x-10=0$

    $⇔x^2-2x+5x-10=0$

    $⇔x.(x-2)+5.(x-2)=0$

    $⇔(x-2).(x+5)=0$

    \(\left[ \begin{array}{l}x-2=0\\x+5=0\end{array} \right.\) 

    \(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\) 

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