a) ĐKXĐ: \(x\ge-4\)
a) Ta có: \(\sqrt{6-4x+x^2}=x+4\Rightarrow\left(x+4\right)^2=x^2-4x+6\)
\(\Rightarrow x^2+8x+16=x^2-4x+6\Rightarrow4x+10=0\Rightarrow x=-\frac{5}{2}\left(loại\right)\)
Vậy pt vô nghiệm
b) \(\sqrt{4x^2-4x+1}+\sqrt{2x-1}=0\Rightarrow\sqrt{\left(2x-1\right)^2}+\sqrt{2x-1}=0\)
\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{2x-1}+1\right)=0\Rightarrow x=\frac{1}{2}\)