Bài 11: Chia đa thức cho đơn thức

a) (x + 2)³ + 1

= (x + 2)³ + 1³

= [(x + 2) + 1][(x + 2)² - (x + 2).1 + 1²]

= (x + 3)(x² + 4x + 4 - x - 2 + 1)

= (x + 3)(x² + 3x + 3)

b) x³ + 6x² + 12x + 9

= x³ + 3x² + 3x² + 9x + 3x + 9

= (x³ + 3x²) + (3x² + 9x) + (3x + 9)

= x²(x + 3) + 3x(x + 3) + 3(x + 3)

= (x + 3)(x² + 3x + 3)

c) x³ + 6x² + 12x + 7

= x³ + x² + 5x² + 5x + 7x + 7

= (x³ + x²) + (5x² + 5x) + (7x + 7)

= x²(x + 1) + 5x(x + 1) + 7(x + 1)

= (x + 1)(x² + 5x + 7)

d) 2x³ + 6x² + 12x + 8

= 2(x³ + 3x² + 6x + 4)

= 2(x³ + x² + 2x² + 2x + 4x + 4)

= 2[(x³ + x²) + (2x² + 2x) + (4x + 4)]

= 2[x²(x + 1) + 2x(x + 1) + 4(x + 1)]

= 2(x + 1)(x² + 2x + 4)

a: =(x+2+1)(x^2+4x+4-x-2+1)

=(x+3)(x^2+3x+3)

b: =x^3+6x^2+12x+8+1

=(x+2)^3+1

=(x+3)(x^2+3x+3)

c: =x^3+x^2+5x^2+5x+7x+7

=(x+1)(x^2+5x+7)

Đây ạ  

1: =x^2-2*x*2y+4y^2

=(x-2y)^2

2: 9x^2-y^2=(3x-y)(3x+y)

5: x^3-8=(x-2)(x^2+2x+4)

4: =x^2-(y-2)^2

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

7: 8x^3+12x^2+6x+1=(2x+1)^3

14: =x^3-3x^2+3x-1-1

=(x-1)^3-1

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

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

15: =x^3+x^2-4x^2-4x+7x+7

=(x+1)(x^2-4x+7)

16: =2x^3-x^2-2x^2+x+2x-1

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

  ạ

14

\(=\left(x-3\right)^3+3^3=\left(x-3+3\right)\left[\left(x-3\right)^2-\left(x-3\right).3+3^2\right]\\ =x\left(x^2-6x+9-3x+9+9\right)=x\left(x^2-9x+27\right)\)

3: =-(x^2-10x+25)

=-(x-5)^2

5: 16-x^2=(4-x)(4+x)

6: 16-(3x+1)^2

=(4-3x-1)(4+3x+1)

=(3x+5)*3(1-x)

7: =(2x+5+3x)(2x+5-3x)

=(5-x)*5*(x+10

8: =(2x-1-3x+1)(2x-1+3x-1)

=-x(5x-2)

9: =(2x-y)^2

10: =(x+1+3y)(x+1-3y)

12: =(y-2)^2-x^2

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

  đây a 

Yêu cầu của đề là gì vậy ?

11: =(x^2y^2)^2+2*x^2y^2*2+2

=(x^2y^2+2)^2

12: =(y-2)^2-x^2

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

13: =(1-3x)(1+3x+9x^2)

14: =(x-3+3)(x^2-6x+9-3x+9+9)

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

15: =(3x+1)^3

16: =(1/3x^2-y)^3

17: =(2x-1+2)(4x^2-4x+1-4x+2+4)

=(2x+1)(4x^2-8x+7)

18: =(2x-1)^3

19: =(2x-1)^3-1

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

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

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

a: =2(x-y)^3/(x-y)-7(x-y)^2/(x-y)+(x-y)/(x-y)

=2(x-y)^2-7(x-y)+1

b: =3(x-y)^5/5(x-y)^2-2(x-y)^4/5(x-y)^2+3(x-y)^2/5(x-y)^2

=3/5(x-y)^3-2/5(x-y)^2+3/5

\(a,\)

\(\left[2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right]:\left(x-y\right)\)

\(=\left[2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right]:\left(x-y\right)\)

\(=\left\{\left(x-y\right)\left[2\left(x-y\right)^2-7\left(x-y\right)+1\right]\right\}:\left(x-y\right)\)

\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)

\(b,\)

\(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:\left[5\left(x-y\right)^2\right]\)

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

Lời giải:
Gọi đa thức dư khi lấy $f(x)$ chia cho $x^2+x-6$ là $ax+b$ với $a,b\in\mathbb{R}$, $Q(x)$ là đa thức thương.

Theo bài ra ta có:

$f(2)=6067$

$f(-3)=-4043$

$f(x)=(x^2+x-6)Q(x)+ax+b=(x-2)(x+3)Q(x)+ax+b$

Cho $x=2$ thì:

$f(2)=0.Q(2)+2a+b=2a+b$

$\Leftrightarrow 6067=2a+b(1)$

Cho $x=-3$ thì:

$f(-3)=0.Q(-3)-3a+b=-3a+b$

$\Leftrightarrow -4043=-3a+b(2)$

Từ $(1); (2)\Rightarrow a=2022; b=2023$

Vậy đa thức dư là $2022x+2023$

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

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

\(\left(10x^2y-8xy^3\right):2xy=5x-4y^2\)

(10x²y - 8xy³) : 2xy

= (10x²y : 2xy) - (8xy³ : 2xy)

= 5x - 4y²

\(1,\left(10x^2y-8xy^3\right):2xy=5x-4y^2\)

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

\(\left(x-y-1\right).\left(x-1+y\right):\left(x-y-1\right)=x-1+y\)

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