\(a,A=\dfrac{\left(6x^2+3x\right)\left(2x-1\right)}{4x^2-1}=\dfrac{3x\left(2x+1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=3x\\ b,A=\dfrac{\left(4x^2-3x-7\right)\left(2x+3\right)}{4x-7}=\dfrac{\left(x+1\right)\left(4x-7\right)\left(2x+3\right)}{4x-7}=\left(x+1\right)\left(2x+3\right)\\ c,A=\dfrac{\left(4x^2-7x+3\right)\left(x^2+2x+1\right)}{x^2-1}=\dfrac{\left(x-1\right)\left(4x-3\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\left(4x-3\right)\left(x+1\right)\\ d,A=\dfrac{\left(2x^2-3x-2\right)\left(x^2+2x\right)}{x^2-2x}=\dfrac{x\left(x+2\right)\left(x-2\right)\left(2x+1\right)}{x\left(x-2\right)}=\left(x+2\right)\left(2x+1\right)\)
