Question

giải phương trình :

\(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)

Answer 1

ĐKXĐ: \(\left[{}\begin{matrix}-1\le x< 0\\x\ge\sqrt{\dfrac{5}{2}}\end{matrix}\right.\)

\(x-\dfrac{4}{x}+\sqrt{2x-\dfrac{5}{x}}-\sqrt{x-\dfrac{1}{x}}=0\)

\(\Leftrightarrow x-\dfrac{4}{x}+\dfrac{x-\dfrac{4}{x}}{\sqrt{2x-\dfrac{5}{x}}+\sqrt{x-\dfrac{1}{x}}}=0\)

\(\Leftrightarrow\left(x-\dfrac{4}{x}\right)\left(1+\dfrac{1}{\sqrt{2x-\dfrac{5}{x}}+\sqrt{x-\dfrac{1}{x}}}\right)=0\)

\(\Leftrightarrow x-\dfrac{4}{x}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\left(loại\right)\end{matrix}\right.\)