Question

Giải phương trình

\(\sqrt{\frac{2x-3}{x-1}}\) = 2

Answer 1

Answer 2

ĐKXĐ : \(x< 1\)

\(\left(\sqrt{\frac{2x-3}{x-1}}\right)^2=2^2\)

=> \(\left|\frac{2x-3}{x-1}\right|=4\)

=> \(\left[{}\begin{matrix}\frac{2x-3}{x-1}=4\\\frac{2x-3}{x-1}=-4\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-3=4\left(x-1\right)\\2x-3=-4\left(x-1\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-3=4x-4\\2x-3=-4x+4\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-4x=3-4\\2x+4x=3+4\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}-2x=-1\\6x=7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-1:\left(-2\right)=\frac{1}{2}=0,5\left(lấy\right)\\x=7:6=\frac{7}{6}=1,1\left(6\right)\left(loại\right)\end{matrix}\right.\)

Vậy \(x\in\left\{0,5\right\}\)