a)\(P=4x^3-\left(2-4x\right).\left(x^2-3x+1\right)\)

\(=4x^3-\left(2x^2-6x+1-4x^2+12x^2-4x\right)\)

\(=4x^3-2x^2+6x-1+4x^2-12x^2+4x\)

\(=4x^3-10x^2+10x-1\)

b) Thay \(x=\frac{-1}{2}\) vào biểu thức trên

Ta Có : \(4.\left(\frac{-1}{2}\right)^3-10.\left(\frac{-1}{2}\right)^2+10.\left(\frac{-1}{2}\right)-1\)

       \(=\frac{-1}{2}-\frac{5}{2}-5-1\)

       \(=-3-5-1\)

\(=-8-1=-9\)

thanks  bạn ạ

Ko có gì

(x +1)³(x-1)+x³-3x(x+1)(x-1)

=(x³+3x²+3x+1)(x-1)+x³-3x(x²-1)

=x⁴-x³+3x³-3x²+3x²-3x+x-1+x³-3x³+3x

=x⁴+x-1

\(=\dfrac{3x^2-x+3-x^2+2x-1-2x^2-2x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{-x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-1}{x^2+x+1}\)

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1-3x^3-3x\)

\(=-x^3+3x\)

\(\left(x-1\right)^3+3\left(x-1\right)^2\cdot x+3\left(x-1\right)\cdot x^2+x^3\)

\(=\left(x-1+x\right)^3\)

\(=\left(2x-1\right)^3\)

(\(3+\dfrac{x}{3-x}+\dfrac{2x}{3+x}-\dfrac{4x^2-3x-9}{x^2-9}\) ):\(\left(\dfrac{2}{3-x}-\dfrac{x-1}{3x-x^2}\right)\)\(=\left(\dfrac{3x^2-27}{\left(x-3\right)\left(x+3\right)}+\dfrac{-x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2-3x-9}{\left(x-3\right)\left(x+3\right)}\right)\)\(:\left(\dfrac{2x}{x\left(3-x\right)}-\dfrac{x-1}{x\left(3-x\right)}\right)\)

\(=\dfrac{3x^2-27-x^2-3x+2x^2-6x-4x^2+3x+9}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{-6x-18}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) \(=\dfrac{-6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{6}{3-x}.\dfrac{x\left(x-3\right)}{x+1}\) \(=\dfrac{6x}{x+1}\)

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)