Question

giải phương trình : \(8x^3-6x=\sqrt{2x+2}\)

Answer 1

\(Đk:x\ge-1\)

\(pt\Leftrightarrow16x^3-12x-2\sqrt{2x+2}=0\Leftrightarrow16x^3-14x-2+\sqrt{2x+2}\left(\sqrt{2x+2}-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(16x^2+16x+2\right)+\sqrt{2x+2}\left(\sqrt{2x+2}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{2x+2}-2\right)\left(\sqrt{2x+2}+2\right)\left(16x^2+16x+2\right)+2\sqrt{2x+2}\left(\sqrt{2x+2}-2\right)=0\)\(\Leftrightarrow\left(\sqrt{2x+2}-2\right)\left(\left(\sqrt{2x+2}+2\right)\left(16x^2+16x+2\right)+2\sqrt{2x+2}\right)=0\)

\(\Leftrightarrow x=1\)