b: Ta có: \(\dfrac{3x-1}{2}=\dfrac{2x-2}{3}\)
\(\Leftrightarrow9x-3=4x-4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\)
\(a,\Leftrightarrow\dfrac{2}{3}\left(3x-2\right)+\dfrac{4}{9}\left(3x-5\right)=-2x+1\\ \Leftrightarrow2x-\dfrac{4}{3}+\dfrac{4}{3}x-\dfrac{20}{9}+2x-1=0\\ \Leftrightarrow\dfrac{16}{3}x=\dfrac{41}{9}\Leftrightarrow x=\dfrac{41}{9}\cdot\dfrac{3}{16}=\dfrac{41}{48}\\ b,\Leftrightarrow3\left(3x-1\right)=2\left(2x-2\right)\\ \Leftrightarrow9x-3=4x-4\\ \Leftrightarrow5x=-1\Leftrightarrow x=-\dfrac{1}{5}\\ c,\Leftrightarrow15-2\left(x^2+1\right)=7\\ \Leftrightarrow13-2x^2=7\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)
\(d,\Leftrightarrow x+1=\sqrt{x+1}\left(x\ge-1\right)\\ \Leftrightarrow x^2+2x+1=x+1\\ \Leftrightarrow x^2-x=0\\ \Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)