Question

giả pt: \(\sqrt{\frac{1-x}{x}}=\frac{2x+x^2}{1+x^2}\)

Answer 1

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

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

\(\Leftrightarrow\frac{-\left(2x-1\right)}{\sqrt{\frac{1-x}{x}}+1}-\frac{2x-1}{1+x^2}=0\)

\(\Leftrightarrow\left(2x-1\right)\left(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}\right)=0\)

Dễ thấy: \(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}< 0\)

\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)