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}}`.
