\(...=A=x^3-3x^2+3x-1+1013\)

\(A=\left(x-1\right)^3+1013=\left(11-1\right)^3+1013=1000+1013=2013\)

\(...B=x^3-6x^2+12x-8-100\)

\(B=\left(x-2\right)^3-100=\left(12-2\right)^3-100=1000-100=900\)

\(...C=\left(x-2y\right)^3=\left(-2y-2y\right)^3=\left(-4y\right)^3=-64y^3\)

\(...D=x^3+9x^2+27x+9+2018\)

\(D=\left(x+3\right)^3+2018=\left(-23+3\right)^3+2018=-8000+2018=-5982\)

a) \(A=x^3-3x^2+3x+1012\)

\(A=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1+1013\)

\(A=\left(x-1\right)^3+1013\)

Thay x=11 vào A ta có:

\(A=\left(11-1\right)^3+1013=10^3+1013=1000+1013=2013\)

b) \(B=x^3-6x^2+12x-108\)

\(B=x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-8-100\)

\(B=\left(x-2\right)^3-100\)

Thay x=12 vào B ta có:

\(B=\left(12-2\right)^3-100=10^3-100=1000-100=900\)

c) \(C=x^3+6x^2y+12xy^2+8y^3\)

\(C=x^3+3\cdot2y\cdot x^2+3\cdot\left(2y\right)^2\cdot x+\left(2y\right)^3\)

\(C=\left(x+2y\right)^3\)

Thay x=-2y vào C ta được:

\(C=\left(-2y+2y\right)^3=0^3=0\)

d) \(D=x^3+9x^2+27x+2027\)

\(D=x^3+3\cdot3\cdot x^2+3\cdot3^2\cdot x+27+2000\)

\(D=\left(x+3\right)^3+2000\)

Thay x=-23 vào D ta có:

\(D=\left(-23+3\right)^3+2000=\left(-20\right)^3+2000=-8000+2000=-6000\)

các bạn giúp mình vs nào

tôi ns thật: bt chết liền

A = x³ - 6x² + 12x - 8

A = x³ - 3x².2 + 3x.2² - 2³

A = (x - 2)³ (1)

Thay x = 22 vào (1) ta có:

A = (22 - 2)³ = 8000

\(B=x^3+3x^2+3x+10.\)

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

\(B=\left(x+1\right)^3+9\)

có x= 99

\(\Rightarrow B=\left(99+1\right)^3+9\)

\(\Rightarrow B=100^3+9=1000009\)

\(\Rightarrow B=x^3+3x^2+3x+10=1000009\)

A=x³-6x²+12x-8

  =x³ - 3.x² .2 + 3x2² - 2³

  = (x-2)³

 - Ta thay x=22 vào biểu thức A thì ta có :

A= (x-2)³ = ( 22-2 )³

A= 20³ = 8000

Vậy khi thay x=22 vào biểu thức A ta được giá trị là 8000

B=x³ +3x²+3x+10 

  = x³ +3x²+3x +1 +9

= (x +1)³+9

- Ta thay x=99 tại biểu thức B ta có :

B= (x+1)³ +9

  = (99+1)³ +9

= 100³ +9

=1000000+9

=1000009

Vậy khi thay x=99 vào biểu thức B ta được giá trị là 100009

\(a,-x^3+3x^2-3x+1=-\left(x^3-3x^2+3x-1\right)=-\left(x^3-3.x^2.1+3.x.1^2-1^3\right)\)

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

\(b,8-12x+6x^2-x^3=2^3-3.2^2.x+3.2.x^2-x^3=\left(2-x\right)^3\)

\(a,x^3+12x^2+48x+64=x^3+3.x^2.4+3.x.4^2+4^3=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

\(b,x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

A = x³ + 3x² + 3x - 899

= (x³ + 3x² + 3x + 1) - 900

= (x + 1)³ - 900

= (29 + 1)³ - 900 = 30³ - 900 = 26100

B = x³ - 6x² + 12x + 10

= (x³ - 6x² + 12x - 8) + 18

= (x - 2)³ + 18

= (12 - 2)³ + 18 = 10³ + 18 = 1000 + 18 = 1018

c) C = 8x³ - 27y³

= (2x)³ - (3y)³

= (2x - 3y)(4x² + 6xy + 9y²)

= (2x - 3y)(4x² - 12xy + 9y²) + (2x - 3y).18xy

= (2x - 3y)(2x - 3y)² + (2x - 3y).18xy

= (2x - 3y)³ + (2x - 3y).18xy

= 5³ + 5.18.4

= 125 - 360

= -235

D = x³ + y³ + 3xy(x² + y²) + 6x²y²(x + y)

= (x + y)(x² - xy + y²) + 3x³y + 3xy³ + 6x²y²

= x² + y² - xy + 3x³y + 3xy³ + 6x²y² 

= (x + y)² - 3xy + 3x³y + 3xy³ + 6x²y² 

= 1 - 3xy(2xy - 1) + 3xy(x² + y²)

= 1 - 3xy(x² + y² + 2xy - 1)

= 1 - 3xy[(x + y)² - 1]

= 1 - 0 = 1

\(a,=\left(x+2\right)^3=100^3=1000000\\ b,=\left(x-3\right)^3=10^3=1000\\ c,Sửa:x^3-3x^2+3x-1=\left(x-1\right)^3=10^3=1000\\ d,Sửa:x^3+15x^2+75x+125=\left(x+5\right)^3=20^3=8000\)

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)