Question

Giải pt: 2.(1+\(\sqrt{ }\)x² +x+1)= x.(x+1)

Answer 1

\(x=\left\{0;-1\right\}\) không phải là nghiệm

Với \(x\ne\left\{0;-1\right\}\)

\(\Leftrightarrow2\left(\sqrt{x^2+x+1}+1\right)\left(\sqrt{x^2+x+1}-1\right)=\left(x^2+x\right)\left(\sqrt{x^2+x+1}-1\right)\)

\(\Leftrightarrow2\left(x^2+x\right)=\left(x^2+x\right)\left(\sqrt{x^2+x+1}-1\right)\)

\(\Leftrightarrow2=\sqrt{x^2+x+1}-1\)

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

\(\Leftrightarrow x^2+x-8=0\Rightarrow x=\frac{-1\pm\sqrt{33}}{2}\)