Question

4x² + 12x\(\sqrt{x+1}\) = 27(x+1)

Answer 1

ĐKXĐ: \(x\ge-1\)

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

\(\Leftrightarrow4a^2+12ab=27b^2\Leftrightarrow4a^2+12ab-27b^2=0\)

\(\Leftrightarrow\left(2a+9b\right)\left(2a-3b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}9b=-2a\\2a=3b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}9\sqrt{x+1}=-2x\left(x\le0\right)\\3\sqrt{x+1}=2x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}81\left(x+1\right)=4x^2\\9\left(x+1\right)=4x^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x^2-81x-81=0\\4x^2-9x-9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{81-9\sqrt{97}}{8}\\x=3\end{matrix}\right.\)