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

(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

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

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

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

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

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

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

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

a: Đặt A=3(5x-2)-|x-5|

TH1: x>=5

=>x-5>=0

A=3(5x-2)-|x-5|

=15x-6-(x-5)

=15x-6-x+5

=14x-1

TH2: x<5

=>x-5<0

A=3(5x-2)-|x-5|

=15x-6-(5-x)

=15x-6-5+x

=16x-11

b: Đặt B=|2x+3|+2x+7

TH1: x>=-3/2

=>2x+3>=0

B=|2x+3|+2x+7

=2x+3+2x+7

=4x+10

TH2: x<-3/2

=>2x+3<0

B=|2x+3|+2x+7

=-2x-3+2x+7

=4

c: đặt C=3x-1+|1-3x|

=3x-1+|3x-1|

TH1: x>=1/3

=>3x-1>=0

C=3x-1+|3x-1|

=3x-1+3x-1

=6x-2

TH2: x<1/3

=>3x-1<0

C=3x-1+|3x-1|

=3x-1+1-3x

=0

d: Đặt D=3(x-1)-2|x+3|

TH1: x>=-3

=>x+3>=0

D=3(x-1)-2|x+3|

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

=3x-3-2x-6

=x-9

TH2: x<-3

=>x+3<0

D=3(x-1)-2|x+3|

=3x-3-2(-x-3)

=3x-3+2x+6

=5x+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: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

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

\(=27x-55\)