Question

Tìm x để thỏa mãn điều kiện  \(\sqrt{\frac{2x-3}{x-1}}=2\)

Answer 1

\(\sqrt{\frac{2x-3}{x-1}}=2\Leftrightarrow\sqrt{\frac{2x-3}{x-1}}=\sqrt{4}\Leftrightarrow\frac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4\cdot\left(x-1\right)\)

\(\Leftrightarrow2x-3=4x-4\)

\(\Leftrightarrow2x-4x=-4+3\)

\(\Leftrightarrow-2x=-1\Leftrightarrow x=\frac{1}{2}\)

Answer 2

ĐKXĐ:

x-1>0

x>1

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

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

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

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

<=>\(2x-3=4.\left(x-1\right)\)

<=>\(2x-3=4x-4\)

<=>\(-2x=-1\)

<=>\(x=\frac{1}{2}\)