a)\(\sqrt{4x^2-4x+1}-\sqrt{9x^2}=0\Leftrightarrow\sqrt{\left(2x-1\right)^2}-\sqrt{\left(3x\right)^2}=0\Leftrightarrow\left|2x-1\right|-\left|3x\right|=0\)TH1: x<0
\(\left|2x-1\right|-\left|3x\right|=0\Leftrightarrow1-2x+3x=0\Leftrightarrow x=-1\)(nhận)
TH2: \(0\le x< \dfrac{1}{2}\)
\(\left|2x-1\right|-\left|3x\right|=0\Leftrightarrow1-2x-3x=0\Leftrightarrow5x=1\Leftrightarrow x=\dfrac{1}{5}\)(nhận)
TH3: \(x\ge\dfrac{1}{2}\)
\(\left|2x-1\right|-\left|3x\right|=0\Leftrightarrow2x-1-3x=0\Leftrightarrow-x=1\Leftrightarrow x=-1\)(loại)
Vậy \(S=\left\{-1;\dfrac{1}{5}\right\}\)
b) \(\sqrt{x^2-2x+1}-\sqrt{3+2\sqrt{2}}=0\Leftrightarrow\sqrt{\left(x-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}=0\Leftrightarrow\left|x-1\right|-\sqrt{2}-1=0\)TH1: x<1
\(\left|x-1\right|-\sqrt{2}-1=0\Leftrightarrow1-x-\sqrt{2}-1=0\Leftrightarrow-x=\sqrt{2}\Leftrightarrow x=-\sqrt{2}\)(nhận)
TH2: x≥1
\(\left|x-1\right|-\sqrt{2}-1=0\Leftrightarrow x-1-\sqrt{2}-1=0\Leftrightarrow x=2+\sqrt{2}\)(nhận)
Vậy: \(S=\left\{-\sqrt{2};2+\sqrt{2}\right\}\)
