a/ ĐKXĐ: \(-4\le x\le\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x+4}=\sqrt{1-x}+\sqrt{1-2x}\)
\(\Leftrightarrow x+4=1-x+1-2x+2\sqrt{2x^2-3x+1}\)
\(\Leftrightarrow2x+1=\sqrt{2x^2-3x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow4x^2+4x+1=2x^2-3x+1\)
\(\Leftrightarrow2x^2+7x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\frac{7}{2}\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\sqrt{3x-2}=\sqrt{x+7}+1\)
\(\Leftrightarrow3x-2=x+8+2\sqrt{x+7}\)
\(\Leftrightarrow x-5=\sqrt{x+7}\) (\(x\ge5\))
\(\Leftrightarrow x^2-10x+25=x+7\)
\(\Leftrightarrow x^2-11x+18=0\Rightarrow\left[{}\begin{matrix}x=9\\x=2\left(l\right)\end{matrix}\right.\)
c/
\(\Leftrightarrow2\left(x^2+1\right)-\left(4x-1\right)\sqrt{x^2+1}+2x-1=0\)
Đặt \(\sqrt{x^2+1}=t\ge1\)
\(2t^2-\left(4x-1\right)t+2x-1=0\)
\(\Delta=\left(4x-1\right)^2-8\left(2x-1\right)=16x^2-24x+9=\left(4x-3\right)^2\)
Phương trình có 2 nghiệm: \(\left[{}\begin{matrix}t=\frac{4x-1-\left(4x-3\right)}{4}=\frac{1}{2}\left(l\right)\\t=\frac{4x-1+4x-3}{4}=2x-1\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+1}=2x-1\) (\(x\ge\frac{1}{2}\))
\(\Leftrightarrow x^2+1=4x^2-4x+1\)
\(\Leftrightarrow3x^2-4x=0\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)
d/ ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x^2-4+1-\sqrt{x-1}+2-\sqrt{2x}=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\frac{2-x}{1+\sqrt{x-1}}+\frac{2\left(2-x\right)}{2+\sqrt{2x}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-\frac{1}{1+\sqrt{x-1}}-\frac{2}{2+\sqrt{2x}}\right)=0\)
\(\Leftrightarrow x=2\)
(do \(x\ge1\Rightarrow\left\{{}\begin{matrix}x+2\ge3\\\frac{1}{1+\sqrt{x-1}}\le1\\\frac{2}{2+\sqrt{2x}}< 1\end{matrix}\right.\) \(\Rightarrow\) ngoặc phía sau luôn dương)
