Bài 2:
\(a,\Leftrightarrow x^2-3x+2=x^2-4x+4\left(x\ge2\right)\\ \Leftrightarrow x=2\left(tm\right)\\ b,\Leftrightarrow\sqrt{x^2-4x+1}=3x-1\\ \Leftrightarrow x^2-4x+1=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\ \Leftrightarrow4x^2-1=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(x\ge\dfrac{1}{3}\right)\\ c,\Leftrightarrow x+2=x^2-6x+9\left(x\ge3\right)\\ \Leftrightarrow x^2-7x+7=0\\ \Leftrightarrow x=\dfrac{7+\sqrt{21}}{2}\left(x\ge3\right)\)
\(d,\Leftrightarrow x^2-x+1=3x+1\left(x\ge-\dfrac{1}{3}\right)\\ \Leftrightarrow x^2-4x=0\\ \Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\\ e,\Leftrightarrow x^2+2x+3=4x+3\left(x\ge-\dfrac{3}{4}\right)\\ \Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(f,\Leftrightarrow\sqrt{3x+2}=\sqrt{x^2+2}\Leftrightarrow3x+2=x^2+2\left(x\ge-\dfrac{2}{3}\right)\\ \Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\\ g,\Leftrightarrow x+2+2\sqrt{x+1}=4-x\left(x\le4\right)\\ \Leftrightarrow\sqrt{x+1}=1-x\\ \Leftrightarrow x+1=x^2-2x+1\left(x\le1\right)\\ \Leftrightarrow x\left(x-3\right)=0\Leftrightarrow x=0\left(x\le1\right)\)
