Question

\(\sqrt{\dfrac{2x-3}{x-1}}\)= 2 . Tìm x

Answer 1

đkxđ: \(\left\{\begin{matrix} 2x-3>=0\\x-1>=0 \end{matrix}\right.\) <=>\(\left\{\begin{matrix} x>=3/2\\x>=1 \end{matrix}\right.\) <=> x>=3/2

hoặc \(\left\{\begin{matrix}2x-3<0\\x-1<0 \end{matrix}\right.\) <=>\(\left\{\begin{matrix} x<3/2\\x<1 \end{matrix}\right.\) <=> x<1

Ta có: \(\sqrt{\dfrac{2x-3}{x-1}}\) =2

<=> \(\dfrac{2x-3}{x-1}\) =4

<=> 2x-3=4x-4

<=> -2x=-1

<=> x=1/2 (T/M)