Tìm x biết: $\frac{2x-1}{7}$ = $\frac{9}{2x-1}$

Đáp án: $x$ $∈$ `{\frac{\sqrt{63}+1}{2};\frac{-\sqrt{63}+1}{2}}`.

Giải thích các bước giải:

`{2x-1}/7 = 9/{2x-1}`

`⇔ (2x-1).(2x-1) = 7.9`

`⇔ (2x-1)^2 = 63`

`⇔ 2x-1 = ±\sqrt{63}`

`⇒` \(\left[ \begin{array}{l}2x=\sqrt{63}+1\\2x=-\sqrt{63}+1\end{array} \right.\) 

$⇒$ \(\left[ \begin{array}{l}x=\dfrac{\sqrt{63}+1}{2}\\x=\dfrac{-\sqrt{63}+1}{2}\end{array} \right.\) 

   Vậy $x$ $∈$ `{\frac{\sqrt{63}+1}{2};\frac{-\sqrt{63}+1}{2}}`.