\(a,VT=\dfrac{x^2y^3\cdot7xy}{5\cdot7xy}=\dfrac{7x^3y^4}{35xy}=VP\\ b,VT=\dfrac{x^2\left(x+2\right):x\left(x+2\right)}{x\left(x+2\right)^2:x\left(x+2\right)}=\dfrac{x}{x+2}=VP\\ c,VT=\dfrac{\left(3-x\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}=\dfrac{\left(x-3\right)^2}{9-x^2}=VP\\ d,VT=\dfrac{x\left(x^2-4\right)}{5\left(2-x\right)}=\dfrac{x\left(x-2\right)\left(x+2\right)}{-5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}=VP\\ e,VT=\dfrac{5y\cdot4x}{7\cdot4x}=\dfrac{20xy}{28x}=VP\\ f,VP=\dfrac{3x\cdot\left(x+5\right)}{2\cdot\left(x+5\right)}=VP\\ g,VP=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+2}{x-1}=VT\)
\(h,VT=\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=x-2=\dfrac{\left(x-2\right)\left(x-1\right)}{x-1}=\dfrac{x^2-3x+2}{x-1}=VP\\ i,VT=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)
