\(a,3\left(x+4\right)=x+2\\ \Leftrightarrow3x+12-x-2=0\\ \Leftrightarrow2x+10=0\\ \Leftrightarrow x=-5\\ b,\dfrac{3x+2}{3}=\dfrac{x}{2}-\dfrac{1}{6}\\ \Leftrightarrow\dfrac{2\left(3x+2\right)}{6}-\dfrac{3x}{6}-\dfrac{1}{6}=0\\ \Leftrightarrow6x+4-3x-1=0\\ \Leftrightarrow3x+3=0\\ \Leftrightarrow x=-1\\ c,\left(2x-1\right)^2=\left(2-x\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(2-x\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2+x\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(3x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
\(d,ĐKXĐ:x\ne0,x\ne-1\\ \dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}-\dfrac{2x-1}{x\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-1+x-2x+1}{x\left(x+1\right)}=0\\ \Rightarrow x^2-x=0\\ \Leftrightarrow x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
