ĐK : tự tìm nha
\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(\Leftrightarrow\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}=x-\frac{4}{x}\)
Đặt \(\sqrt{x-\frac{1}{x}}=a;\sqrt{2x-\frac{5}{x}}=b\left(a,b\ge0\right)\Rightarrow x-\frac{1}{x}=a^2;2x-\frac{5}{x}=b^2\)
\(\Rightarrow b^2-a^2=x-\frac{4}{x}\)
Ta có : \(a-b=b^2-a^2\)
\(\Leftrightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(1+a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{x}-x=0\\a+b=-1\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow4-x^2=0\Leftrightarrow x=2\) ( t/m )
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