Question

Gải phương trình:

\(x^2+2x\sqrt{x+\frac{1}{x}}=8x-1\)

 

Answer 1

ĐK: \(x\ne0;x+\frac{1}{x}=\frac{x^2+1}{x}\ge0\Leftrightarrow x>0\)

\(pt\Leftrightarrow x^2-4x+1+2x\left(\sqrt{\frac{x^2+1}{x}}-2\right)=0\)

\(\Leftrightarrow\left(x^2-4x+1\right)+2x.\frac{\frac{x^2+1}{x}-4}{\sqrt{x+\frac{1}{x}}+2}=0\)

\(\Leftrightarrow\left(x^2-4x+1\right)+2.\frac{x^2-4x+1}{\sqrt{x+\frac{1}{x}}+2}=0\)

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

\(\Leftrightarrow x^2-4x+1=0\text{ (do }1+\frac{2}{\sqrt{x+\frac{1}{x}}+2}>0\text{)}\)

\(\Leftrightarrow x=2+\sqrt{3};\text{ }2-\sqrt{3}\text{ (nhận)}\)

Kết luận: \(x=2+\sqrt{3};\text{ }x=2-\sqrt{3}\)