a/ ĐKXĐ: \(x\ge-\frac{1}{3}\)
\(\Leftrightarrow\sqrt{5x+7}=\sqrt{x+3}+\sqrt{3x+1}\)
\(\Leftrightarrow5x+7=4x+4+2\sqrt{\left(x+3\right)\left(3x+1\right)}\)
\(\Leftrightarrow x+3=2\sqrt{3x^2+10x+3}\)
\(\Leftrightarrow x^2+6x+9=4\left(3x^2+10x+3\right)\)
\(\Leftrightarrow11x^2+34x+3=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\x=-\frac{1}{11}\end{matrix}\right.\)
b/ ĐKXĐ: \(-2\le x\le2\)
Đặt \(\sqrt{2-x}+\sqrt{2+x}=t>0\)
\(\Rightarrow t^2=4+2\sqrt{4-x^2}\Rightarrow\sqrt{4-x^2}=\frac{t^2-4}{2}\) (1)
Pt trở thành:
\(\Rightarrow t+\frac{t^2-4}{2}=2\)
\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\left(l\right)\end{matrix}\right.\)
Thế vào (1):
\(\sqrt{4-x^2}=\frac{t^2-4}{2}=0\)
\(\Leftrightarrow4-x^2=0\Rightarrow x=\pm2\)
