Question

Giải các pt sau:

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

Answer 1

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

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

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

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

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

Suy ra x=2 và \(\frac{1}{\sqrt{x+2}+2}=\frac{x}{2x+1}\) *nhân chéo, bình phương rút gọn*

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