Question

[QMing]

Giải phương trình:\(3x^2+3x+2=\left(x+6\right)\sqrt{3x^2-2x-3}\)

Answer 1

[Nguyễn Minh Đăng]

ĐK: \(\orbr{\begin{cases}x\ge\frac{1+\sqrt{10}}{3}\\-6\le x\le\frac{1-\sqrt{10}}{3}\end{cases}}\)

Ta có: \(3x^2+3x+2=\left(x+6\right)\sqrt{3x^2-2x-3}\)

\(\Leftrightarrow\left[\left(3x^2-2x-3\right)-5\sqrt{3x^2-2x-3}\right]+\left[5\left(x+1\right)-\left(x+1\right)\sqrt{3x^2-2x-3}\right]=0\)

\(\Leftrightarrow\left(\sqrt{3x^2-2x-3}-5\right)\sqrt{3x^2-2x-3}-\left(x+1\right)\left(\sqrt{3x^2-2x-3}-5\right)=0\)

\(\Leftrightarrow\left(\sqrt{3x^2-2x-3}-5\right)\left(\sqrt{3x^2-2x-3}-x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{3x^2-2x-3}=5\\\sqrt{3x^2-2x-3}=x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x^2-2x-3=25\\3x^2-2x-3=x^2+2x+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x^2-2x-28=0\\2x^2-4x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1\pm\sqrt{85}}{3}\left(tm\right)\\x=1\pm\sqrt{3}\left(tm\right)\end{cases}}\)