Bài 7

Ta có:

VT = a³ + b³ = (a + b)(a² - ab + b²)

= (a + b)(a² - 2ab + b² + ab)

= (a + b)[(a - b)² + ab]

= VP

Vậy a³ + b³ = (a + b)[(a - b)² + ab]

Bài 8

(x + 2)³ + (x - 2)³ + x³ - 3x(x + 2)(x - 2)

= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 + x³ - 3x(x² - 4)

= 3x³ + 24x - 3x³ + 12x

= 36x

  đây nhé làm 3 bài đầu 

5:

B=(x+y)^2-2xy

=225-2*(-100)=425

6:

a: C=(39+61)^2=100^2=10000

b: D=50^2-50^2+1=1

  bài đây nhé mn nhanh nhanh giúp mik với

a: (x-1/x)^2=x^2-2+1/x^2

b: (1/2x+1/3y)(1/4x^2-1/6xy+1/9y^2)=1/8x^3+1/27y^3

\(\left(5x-10\right)\left(x^2-1\right)-\left(3x-6\right)\left(x^2-2x+1\right)\)

\(=\left(5x-10\right)\left(x-1\right)\left(x+1\right)-\left(3x-6\right)\left(x-1\right)^2\)

\(=\left(x-1\right)\left[\left(5x-10\right)\left(x+1\right)-\left(3x-6\right)\left(x-1\right)\right]\)

\(=\left(x-1\right)\left[5\left(x-2\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)\right]\)

\(=\left(x-1\right)\left[\left(x-2\right)\left(5x+5-3x+3\right)\right]\)

\(=\left(x-1\right)\left[\left(x-2\right)\left(2x+8\right)\right]\)

\(=\left(x-1\right)\left(x-2\right)\left(2x+8\right)\)

Sao mình không thấy gì nhỉ?

a: =6y^3-3y^2-y^2+2y-y+y^2-y^3

=5y^3-3y^2+y

b: =2x^2a-a-2x^2a-a-x^2-ax

=-x^2-ax-2a

c: =2p^3-p^3+1+2p^3+6p^2-3p^5

=3p^3+6p^2-3p^5+1

d: =-3a^3+5a^2+4a^3-4a^2=a^3+a^2

\(a,3\left(2a-1\right)+5\left(3-a\right)\)

\(=6a-3+15-5a\)

\(=a-12\)

Thay \(a=\dfrac{-3}{2}\) vào biểu thức trên 

\(a-12\)

\(=\dfrac{-3}{2}-12\)

\(=\dfrac{-27}{2}\)

\(b,25x-4\left(3x-1\right)+7\left(5-2x\right)\)

\(=25x-12x+4+35-14x\)

\(=-1x+39\)

Thay \(x=2,1\) vào biểu thức trên

\(-1x+39\)

\(=-1.2,1+39\)

\(=-2,1+39\)

\(=36,9\)

\(c,4a-2\left(10a-1\right)+8a-2\)

\(=4a-20a+2+8a-2\)

\(=-8a\)

Thay \(a=-0,2\) vào biểu thức trên

\(-8a\)

\(=-8.\left(-0,2\right)\)

\(=1,6\)

\(d,12\left(2-3b\right)+35b-9\left(b+1\right)\)

\(=24-36b+35b-9b-9\)

\(=-10b-15\)

Thay \(b=\dfrac{1}{2}\) vào biểu thức trên 

\(-10b-15\)

\(=-10.\dfrac{1}{2}-15\)

\(=-20\)

\(a,3x\left(5x^2-2x-1\right)=15x^3-6x^2-3x\)

\(b,\left(x^2-2xy+3\right)\left(-xy\right)=-x^3y+2x^2y^2-3xy\)

\(c,\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)=x^5y-\dfrac{2}{25}x^3y^3-\dfrac{1}{2}x^2y\)

\(d,\dfrac{2}{7}x\left(1,4x-3,5y\right)=\dfrac{2}{5}x^2-xy\)

\(e,\dfrac{1}{2}xy\left(\dfrac{2}{3}x^2-\dfrac{3}{4}x+\dfrac{4}{5}y^2\right)=\dfrac{3}{4}x^3y-\dfrac{3}{8}x^2y+\dfrac{2}{5}xy^3\)

\(f,\left(1+2x-x^2\right)5x=5x+10x^2-5x^3\)

a: =4x+2x^2-x^3-2x^2+x^3-4x+3=3

b: =24-4x+2x^2+3x^3-5x^2+4x+3x^2-3x^3=24

c: =x^4-x^3-3x^2+2x-x^4-x^3-3x^2+2x^2+2x+6+4x^2-4x-8

=-2

d: =x^2n-2x^n+x^n-2-x^2n+x^n+2009=2009