a, Với x > 1
\(P=\dfrac{x^3+4x-3-4\left(x-1\right)-2}{x^2+x+1}=\dfrac{x^3-1}{x^2+x+1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)
b, Với 3 =< x < 13/4
\(Q=\left(x+3\right)\left(x-3-1\right)+\left(13-4x\right)\)
\(=\left(x+3\right)\left(x-4\right)+13-4x=x^2-x-12+13-4x=x^2-5x+1\)

