Bài 1:
\(a,VT=\dfrac{x^2\left(x+2\right):x\left(x+2\right)}{x\left(x+2\right)^2:x\left(x+2\right)}=\dfrac{x}{x+2}=VP\\ b,VT=\dfrac{x\left(x^2-4\right)}{5\left(2-x\right)}=\dfrac{-x\left(2-x\right)\left(x+2\right)}{5\left(x-2\right)}=\dfrac{-x^2-2x}{5}=VP\\ c,VP=\dfrac{2x-1}{\left(2x-1\right)\left(x+2\right)}=\dfrac{1}{x+2}=VT\)
Bài 2:
\(a,A=\dfrac{\left(2x-3\right)\left(2x^2+3x\right)}{4x^2-9}=\dfrac{x\left(2x+3\right)\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=x\\ b,A=\dfrac{\left(2b^2-3b-9\right)\left(b^2+3b\right)}{b^2-3b}=\dfrac{b\left(b+3\right)\left(b-3\right)\left(2b+3\right)}{b\left(b-3\right)}=\left(b+3\right)\left(2b+3\right)\)