Question

Giải phương trình :

\(2\left(x^2+2x+3\right)=5\sqrt{x^3+3x^2+3x+2}\)

Answer 1

ĐKXĐ: \(x\ge-2\)

\(\Leftrightarrow2\left(x^2+2x+3\right)=5\sqrt{\left(x+2\right)\left(x^2+x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\\\sqrt{x^2+x+1}=b\end{matrix}\right.\)

\(\Rightarrow2\left(a^2+b^2\right)=5ab\)

\(\Leftrightarrow2a^2-5ab+2b^2=0\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\sqrt{x^2+x+1}\\2\sqrt{x+1}=\sqrt{x^2+x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\left(x^2+x+1\right)\\4\left(x+1\right)=x^2+x+1\end{matrix}\right.\) \(\Leftrightarrow...\)