Question

giải phương trình:

2(x²+2x+3)=5\(\sqrt{x^3+3x^2+3x+2}\)

Answer 1

ĐKXĐ: \(x\ge-2\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\\\sqrt{x^2+x+1}=b\end{matrix}\right.\) \(\Rightarrow2a^2+2b^2-5ab=0\)

\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+2}=\sqrt{x^2+x+1}\\\sqrt{x+2}=2\sqrt{x^2+x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x-7=0\\4x^2+3x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{37}}{2}\\x=\dfrac{3+\sqrt{37}}{2}\end{matrix}\right.\)