6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)

\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)

7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)

\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)

8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)

\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)

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

10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)

\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)

11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)

12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)

13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)

Yêu cầu đề là gì vậy bạn?

\(b,\left(x+2\right)^2-25\)

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

\(=\left(x-3\right)\left(x+7\right)\)

\(c,36\left(x-y\right)^2\)

\(=36\left(x^2-2xy+y^2\right)\)

\(=36x^2-72xy+36y^2\)

\(d,x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

\(=x^2+2.x.\dfrac{1}{4}+\dfrac{1}{4}^2\)

\(=\left(x+\dfrac{1}{4}\right)^2\)

\(e,2x^4y^3-3x^2y^4+5x^3y^4\)

\(=x^2y^3\left(2x^2-3y+5xy\right)\)

Các câu còn lại làm tương tự, chú ý sd HĐT

Bài 6:

M= 2.2 - 2.3+3.2.3

M= 4 - 6 + 18

M= 20

Bài 7: 

P= 1.2 - 5.-1.-2 + 8.-2.2

P = 2 -10 -32

P= -44

Bài 8:

A (thiếu dữ kiện bn ơi)

B= -1.2 . 3.2 + -1.3 +3.3 +-1.3

B= -2 . 6 + -3 + 9 +-3

B= -2 . 6 - 3 + 9 - 3

B= -12 - 3 + 9 - 3

B= -9

`#3107.101107`

`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`

Ta có:

`x - y - 1 = 0`

`=> x - y = 1`

`D = x^3 - y^3 - 3xy`

`= (x - y)(x^2 + xy + y^2) - 3xy`

`= 1 * (x^2 + xy + y^2) - 3xy`

`= x^2+ xy + y^2 - 3xy`

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

`= x^2 - 2*x*y + y^2`

`= (x - y)^2`

`= 1^2 = 1`

Vậy, với `x - y = 1` thì `D = 1`

________

`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`

`x + y = 5`

`=> (x + y)^2 = 25`

`=> x^2 + 2xy + y^2 = 25`

`=> 2xy = 25 - (x^2 + y^2)`

`=> 2xy = 25 - 17`

`=> 2xy = 8`

`=> xy = 4`

Ta có:

`E = x^3 + y^3`

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

`= 5 * [ (x^2 + y^2) - xy]`

`= 5 * (17 - 4)`

`= 5 * 13`

`= 65`

Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`

________

`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`

Ta có:

`x - y = 4`

`=> (x - y)^2 = 16`

`=> x^2 - 2xy + y^2 = 16`

`=> (x^2 + y^2) - 2xy = 16`

`=> 2xy = (x^2 + y^2) - 16`

`=> 2xy = 26 - 16`

`=> 2xy = 10`

`=> xy = 5`

Ta có:

`F = x^3 - y^3`

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

`= 4 * [ (x^2 + y^2) + xy]`

`= 4 * (26 + 5)`

`= 4*31`

`= 124`

Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`

\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)

 a) Ta thấy \(xy=\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\dfrac{3^2-5}{2}=2\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\) \(=3\left(5-2\right)=9\)

 b) Ta thấy \(xy=\dfrac{-\left(x-y\right)^2+\left(x^2+y^2\right)}{2}=\dfrac{15-5^2}{2}=-5\)

\(\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\) \(=5\left(15-5\right)=50\)

a) \(\left(x^5+4x^3-6x^2\right):4x^2\)

\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)

\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

b) 

Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)

c) (-2x⁵ : 2x²) + (3x² : 2x²) + (-4x^3 : 2x^2)

= \(-x^3+\dfrac{3}{2}-2x\)

d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)

\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)

\(=x-4\)

(dùng hẳng đẳng thức thứ 7)

Bài 2 :

a) 3x(x - 2) - 5x(1 - x) - 8(x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= (3x² + 5x² - 8x²) + (-6x - 5x) + 24 

= -11x + 24

b) (x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³ 

c) (x - y)² + (x + y)² - 2(x - y)(x + y)

= (x - y)² - 2(x - y)(x + y) + (x + y)²

= [(x - y) + x + y)² = [x - y + x + y] = (2x)² = 4x²

Bài 1 :

a]=  \(\frac{1}{4}\)x³ + x - \(\frac{3}{2}\).

b] => [x³ + x² -12 ] = [ x² +3 ][x-2] + [-6]

c]= -x³ -2x +\(\frac{3}{2}\).

d] = [ x³ - 64 ]  = [ x² + 4x + 16][ x- 4].

a) A = -1;                        b) B = ( x   +   y ) 3  =1.