a,\(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)

\(=\left(2x\right)^3-1=8x^3-1\)

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

\(=x^2+2.x.2y+\left(2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)

`a)(2x-1)(4x^2+2x+1)`

`=(2x-1)[(2x)^2+2x.1+1^2]`

`=(2x)^3-1^3`

`=8x^3-1`

Áp dụng HĐT:`A^3-B^3=(A-B)(A^2+AB+B^2)`

`b)(x+2y+z)(x+2y-z)`

`=[(x+2y)+z][(x+2y)-z]`

`=(x+2y)^2-z^2`

`=x^2+2.x.2y+(2y)^2-z^2`

`=x^2+4xy+4y^2-z^2`

Áp dụng HĐT:`A^2-B^2=(A+B)(A-B)`

                      `(A+B)^2=A^2+2AB+B^2`

a) \(\left(2x-1\right)\left(4x^2+2x+1\right)=8x^3-1\)

b) \(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)

a) \(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x\right)^3-1^3=8x^3-1\)

b) \(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2.\)

\(=3x^2\left(x^2-1\right)+\left(x^8-3x^4+3x^2-1\right)-\left(x^8-1\right)\)

\(=3x^4-3x^2+x^8-3x^4+3x^2+1-x^8+1\)

\(=2\)

=2 nha ban

(con cach lam ban nhan dang thuc len rui rut gon lai)

a) = (x+1-x+1)(x²+2x+1+x²-1+x²-2x+1)- 6(x²-1)

   = 2( 3x²+1)- 6(x²-1)

   = 2( 3x²+1-3x²+3)

   =2. 4

   =8

a) \(\left(1+x\right)^2+\left(1-x\right)^2\) 

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

\(=2x^2+2\)

b) \(\left(x+2\right)^2+\left(1+x\right)\left(1-x\right)\)

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

\(=4x+5\)

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

\(=x^2-6x+9+3x^2+6x+3\)

\(=4x^2+12\)

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

\(=9x^2-4-9x^2-6x-1\)

\(=-6x-5\)

e) \(\left(x+5\right)\left(x-2\right)-\left(x+2\right)^2\)

\(=x^2-2x+5x-10-x^2-4x-4\)

\(=-x-14\)

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

\(=2x^2-5x+6x-15-2-4x-2x^2\)

\(=-3x-17\)

g) \(\left(4x-1\right)\left(4x+1\right)-4\left(1-2x\right)^2\)

\(=16x^2-1-4+16x-16x^2\)

\(=16x-5\)

#Học tốt!

a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

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

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

\(=\left(x+1\right)\cdot\left(-1\right)\)

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

b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)

\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)

\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)

\(=x^3-1-x^3-8+12x-12\)

\(=-21+12x\)

c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)

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

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

\(=0\)

câu b bạn làm sai rồi í!

`C=(x+2y)^3-6(x+2y)^2+12(x+2y)-8`

`C=(x+2y-2)^3` (HĐT số `5`)

Thay `x=20;y=1` vào `C` có:

 `C=(20+2.1-2)^3=8000`.

a: \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)

\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)