\(1,\\ a,ĐK:x\ne2\\ b,Đk:\left(x-2\right)^2\ne0\Leftrightarrow x\ne2\\ c,ĐK:-2x+4\ne0\Leftrightarrow x\ne2\\ d,ĐK:x\left(x+2\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\\ 2,\\ a,VT=\dfrac{2\left(x-y\right)}{-3\left(x-y\right)}=-\dfrac{2}{3}=VP\\ b,VP=\dfrac{x^2}{x\left(x+2\right)}=\dfrac{x}{x+2}=VP\\ c,VT+\dfrac{\left(2-3x\right)\left(2+3x\right)}{2\left(2-3x\right)}=\dfrac{2+3x}{2}=VP\\ 3,\\ a,ĐK:x\ne2\\ PT\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,ĐK:x\ne\dfrac{5}{4}\\ PT\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\\ c,ĐK:x\ne1\\ PT\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}=0\\ \Leftrightarrow\dfrac{x+1}{x-1}=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\left(tm\right)\)
\(4,\\ a,ĐK:x^2+1\ne0\left(luôn.đúng.do.x^2+1\ge1>0\right)\\ b,ĐK:x^2+2x+4\ne0\left(luôn.đúng.do.x^2+2x+4=\left(x+1\right)^2+3>0\right)\\ c,ĐK:-x^2+4x-5\ne0\left(luôn.đúng.do.-x^2+4x-5=-\left(x-2\right)^2-1< 0\right)\)
