Question

[Nguyen Duy Dai]

Giải phương trình \(\sqrt{x^2+x+2}=\frac{3x^2+3x+2}{3x+1}\)

Answer 1

[Trí Tiên]

ĐKXĐ : \(x\ne-\frac{1}{3}\)

Ta có : \(\sqrt{x^2+x+2}=\frac{3x^2+3x+2}{3x+1}\)

\(\Leftrightarrow\sqrt{x^2+x+2}-2=\frac{3x^2+3x+2}{3x+1}-2\)

\(\Leftrightarrow\frac{x^2+x+2-4}{\sqrt{x^2+x+2}+2}=\frac{3x^2+3x+2-6x-2}{3x+1}\)

\(\Leftrightarrow\frac{x^2+x-2}{\sqrt{x^2+x+2}+2}=\frac{3x^2-3x}{3x+1}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{3x\left(x-1\right)}{3x+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left[\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{3x}{3x+1}\right]=0\)

\(\Leftrightarrow x=1\)( Thỏa mãn )