\(1,\\ 1,\\ a,=x\left(x-3y\right)\\ b,=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\\ c,=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\\ 2,\\ a,\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ b,\Leftrightarrow3x-x^2+x^2=5\Leftrightarrow3x=5\Leftrightarrow x=\dfrac{5}{3}\\ c,\Leftrightarrow\left(x-2\right)^2-9=0\\ \Leftrightarrow\left(x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\\ 2,\\ a,ĐK:x\ne\pm2\\ b,A=\dfrac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{x-2}\\ b,x=3\Leftrightarrow A=\dfrac{5}{1}=5\\ c,A=1-\dfrac{4}{x-2}\in Z\Leftrightarrow x-2\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow x\in\left\{-2;0;1;3;4;6\right\}\)
\(4,\\ M=\left(x^2+\dfrac{1}{4}y^2+1-xy+2x-y\right)+\dfrac{3}{4}y^2-3y+3-2022\\ M=\left(x-\dfrac{1}{2}y+1\right)^2+\dfrac{3}{4}\left(y^2-4y+4\right)-2022\\ M=\left(x-\dfrac{1}{2}y+1\right)^2+\dfrac{3}{4}\left(y-2\right)^2-2022\ge-2022\\ M_{min}=-2022\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}y-1=0\\y=2\end{matrix}\right.\)
