Question

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

Answer 1

Điều kiện xác định tự làm:

Đặt \(\left\{{}\begin{matrix}\sqrt{x-\dfrac{1}{x}}=a\ge0\\\sqrt{2x-\dfrac{5}{x}}=b\ge0\end{matrix}\right.\)

Ta có

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

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

\(\Leftrightarrow b-a+b^2-a^2=0\)

\(\Leftrightarrow\left(b-a\right)\left(b+a+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b=-1\left(l\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow x^2=4\)

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