Bài 2:
\(a,A=\dfrac{\left(2x-3\right)\left(2x^2+3x\right)}{4x^2-9}=\dfrac{x\left(2x-3\right)\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}=x\\ b,A=\dfrac{\left(2b^2-3b-9\right)\left(b^2+3b\right)}{b^2-3b}=\dfrac{b\left(b-3\right)\left(2b+3\right)\left(b+3\right)}{b\left(b-3\right)}=\left(2b+3\right)\left(b+3\right)\\ c,B=\dfrac{\left(2y-1\right)\left(y^2-4y+3\right)}{y-3}=\dfrac{\left(y-3\right)\left(y-1\right)\left(2y-1\right)}{y-3}=\left(y-1\right)\left(2y-1\right)\\ B=\dfrac{\left(a-1\right)\left(a^3-8\right)}{a^2+2a+4}=\dfrac{\left(a-1\right)\left(a-2\right)\left(a^2+2a+4\right)}{a^2+2a+4}=\left(a-1\right)\left(a-2\right)\)
a: \(\dfrac{2x-1}{2x^2+3x-2}=\dfrac{2x-1}{2x^2+4x-x-2}=\dfrac{1}{x+2}\)
