Ta có: \(1+6x-6x^2-x^3\)

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

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

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

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

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

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

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

=(1-x)(1+x+x^2)+6x(1-x)

=(1-x)( 1+ x+ x^2 + 6x)

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

Đây bạn ơi!

1)

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

2)

\(8x^3-12x^2+6x-2\\ =2\left(4x^3-6x^2+3x-1\right)\\ =2\left(4x^3-4x^2-2x^2+2x+x-1\right)\\ =2\left(4x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\right)\\ =2\left(x-1\right)\left(4x^2-2x+1\right)\)

1) \(y^2-4y+4-x^2\)

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

\(=\left(y-2\right)^2-x^2\)

\(=\left(y-2-x\right)\left(y-2+x\right)\)

2) \(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

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

1: Đa thức này ko phân tích được nha bạn

2: \(x^2+8x+7\)

\(=x^2+x+7x+7\)

\(=x\left(x+1\right)+7\left(x+1\right)\)

\(=\left(x+1\right)\left(x+7\right)\)

3: \(x^2-6x-16\)

\(=x^2-8x+2x-16\)

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

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

4: \(4x^2-8x+3\)

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

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

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

5: \(3x^2-11x+6\)

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

\(=3x\left(x-3\right)-2\left(x-3\right)\)

\(=\left(x-3\right)\left(3x-2\right)\)

1) \(2xy^3-6x^2+10xy\)

\(=2x.y^3-2x.3x+2x.5y\)

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

\(=2x[y\left(y^2-5\right)-3x]\)

2) \(a^6-a^5-2a^3+2a^2\)

\(=\left(a^6-a^5\right)-\left(2a^3-2a^2\right)\)

\(=\left(a^5.a-a^5.1\right)-\left(2a^2.a-2a^2.1\right)\)

\(=a^5\left(a-1\right)-2a^2\left(a-1\right)\)

\(=\left(a^5-2a^2\right)\left(a-1\right)\)

\(=a^2\left(a^3-2\right)\left(a-1\right)\)

3: \(\left(a+b\right)^3-\left(a-b\right)^3\)

\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)

\(=2b\left(3a^2+b^2\right)\)

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

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

1) Ta có: \(x^3+2x^2-6x-27\)

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

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

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

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

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

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

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

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

5: \(x^2+2x-4y^2+1\)

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

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

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

6: \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

\(-9x^2+6x+y^2-1\)

\(=-\left(9x^2-6x+1-y^2\right)\)

\(=-\left(3x-1-y\right)\left(3x-1+y\right)\)

a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)

b.  \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)

c.  \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)

a) $x^3-3x^2y+4x-12y$

$=(x^3-3x^2y)+(4x-12y)$

$=x^2(x-3y)+4(x-3y)$

$=(x-3y)(x^2+4)$

b) $4x^2-y^2+4y-4$

$=4x^2-(y^2-4y+4)$

$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$

$=(2x)^2-(y-2)^2$

$=[2x-(y-2)][2x+(y-2)]$

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

c) $9x^2-6x-y^2+2y$

$=(9x^2-y^2)-(6x-2y)$

$=[(3x)^2-y^2]-2(3x-y)$

$=(3x-y)(3x+y)-2(3x-y)$

$=(3x-y)(3x+y-2)$

$\text{#}Toru$

bạn ấn ở chỗ x² cho rõ hơn nhé