a.
ĐKXĐ: \(x\ge-1\)
\(7+12\sqrt{x+1}=x+4\sqrt{x^2+3x+2}\)
\(\Leftrightarrow4\sqrt{\left(x+1\right)\left(x+2\right)}-12\sqrt{x+1}+x-7=0\)
\(\Leftrightarrow4\sqrt{x+1}\left(\sqrt{x+2}-3\right)+x-7=0\)
\(\Leftrightarrow4\sqrt{x+1}\left(\dfrac{x-7}{\sqrt{x+2}+3}\right)+x-7=0\)
\(\Leftrightarrow\left(x-7\right)\left(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1\right)=0\)
\(\Leftrightarrow x-7=0\) (do \(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1>0;\forall x\ge-1\))
\(\Rightarrow x=7\)
b.
ĐKXĐ: \(x\ne-\dfrac{1}{3}\)
\(\Rightarrow3x^2+3x+2=\left(3x+1\right)\sqrt{x^2+x+2}\)
\(\Leftrightarrow x^2+x+2-\left(3x+1\right)\sqrt{x^2+x+2}+2x^2+2x=0\)
Đặt \(\sqrt{x^2+x+2}=t\)
\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)
\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1+x-1}{2}=2x\\t=\dfrac{3x+1-\left(x-1\right)}{2}=x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2x\left(x\ge0\right)\\\sqrt{x^2+x+2}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4x^2\left(x\ge0\right)\\x^2+x+2=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\\\end{matrix}\right.\)
