c.2
ĐKXĐ: \(x\ge0\)
\(2x^2+5x+4=6\sqrt{2x^3+4x}\)
\(\Leftrightarrow2\left(x^2+2\right)+5x=6\sqrt{2x\left(x^2+2\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x}=a\ge0\\\sqrt{x^2+2}=b>0\end{matrix}\right.\)
\(\Rightarrow2b^2+\dfrac{5a^2}{2}=6ab\)
\(\Leftrightarrow5a^2-12ab+4b^2=0\)
\(\Leftrightarrow\left(5a-2b\right)\left(a-2b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5a=2b\\a=2b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}5\sqrt{2x}=2\sqrt{x^2+2}\\\sqrt{2x}=2\sqrt{x^2+2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}50x=4\left(x^2+2\right)\\2x=4\left(x^2+2\right)\end{matrix}\right.\) (chuyển vế bấm máy)
ĐKXĐ: \(x\ge-3\)
\(\sqrt{x^2-2x+3}=\sqrt{x+3}\)
\(\Leftrightarrow x^2-2x+3=x+3\)
\(\Leftrightarrow x^2-3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (thỏa mãn)
