Tìm x biết : a. (5x-4)^2 – 49x^2 = 0 b. x^2 +3x-10 =0 06/07/2021 Bởi Faith Tìm x biết : a. (5x-4)^2 – 49x^2 = 0 b. x^2 +3x-10 =0
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: $a)(5x-4)^2-49x^2=0$ $⇔(5x-4-7x).(5x-4+7x)=0$ $⇔(-2x-4).(12x-4)=0$ ⇔\(\left[ \begin{array}{l}-2x-4=0\\12x-4=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{3}\end{array} \right.\) $b)x^2+3x-10=0$ $⇔x^2-2x+5x-10=0$ $⇔x.(x-2)+5.(x-2)=0$ $⇔(x-2).(x+5)=0$ \(\left[ \begin{array}{l}x-2=0\\x+5=0\end{array} \right.\) \(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\) Bình luận
`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` \(⇔\left[ \begin{array}{l}x+5=0\\x-2=0\end{array} \right.⇔\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\) Bình luận
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:
$a)(5x-4)^2-49x^2=0$
$⇔(5x-4-7x).(5x-4+7x)=0$
$⇔(-2x-4).(12x-4)=0$
⇔\(\left[ \begin{array}{l}-2x-4=0\\12x-4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{3}\end{array} \right.\)
$b)x^2+3x-10=0$
$⇔x^2-2x+5x-10=0$
$⇔x.(x-2)+5.(x-2)=0$
$⇔(x-2).(x+5)=0$
\(\left[ \begin{array}{l}x-2=0\\x+5=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\)
`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`
\(⇔\left[ \begin{array}{l}x+5=0\\x-2=0\end{array} \right.⇔\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\)