a, đk \(x\ne1\)
Thay \(x=3\) vào A ta được
\(A=\dfrac{4.\left(3+2\right)}{3-1}=\dfrac{4.5}{2}=\dfrac{20}{2}=10 \)
Vậy với \(x=3\) thì \(A=10\)
b, đk \(x\ne-1;x\ne1;x\ne2\)
\(B=\dfrac{\left(x+1\right)^2.\left(x-2\right)-\left(1-x\right)^2.\left(x-2\right)-4x^2}{\left(1-x\right)\left(x+1\right)\left(x-2\right)}\\ =\dfrac{\left(x^2+2x+1\right)\left(x-2\right)-\left(1-2x+x\right)^2\left(x-2\right)-4x^2}{\left(1-x\right)\left(x+1\right)\left(x-2\right)}\\ =\dfrac{x^3+2x^2+x-2x^2-4x-2-\left(x-2x^2+x^3-2+4x-2x^2\right)-4x^2}{\left(1-x\right)\left(x+1\right)\left(x-2\right)}\\ =\dfrac{x^3-3x-2-x^3+4x^2-5x+2-4x^2}{\left(1-x^2\right)\left(x-2\right)}\\ =\dfrac{-8x}{\left(1-x^2\right)\left(x-2\right)}\)
\(P=A:B=\dfrac{4\left(x+2\right)}{x-1}:\dfrac{-8x}{\left(1-x^2\right)\left(x-2\right)}\\ =\dfrac{4\left(x+2\right)}{x-1}.\dfrac{\left(1-x^2\right)\left(x-2\right)}{-8x}\\ =\dfrac{\left(x+2\right)}{x-1}.\dfrac{\left(x^2-1\right)\left(x-2\right)}{2x}\\ =\dfrac{\left(x+2\right)\left(x-2\right)\left(x+1\right)}{2x}\\ =\dfrac{\left(x^2-4\right)\left(x+1\right)}{2x}\)
