`x^3+y^3+21x+21y`

`=(x+y)(x^2-xy+y^2)+21(x+y)`

`=(x+y)(x^2-xy+y^2+21)`

\(x^3+y^3+21x+21y\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+21\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2+21\right)\)

\(l,x^3+y^3+21x+21y\\=(x^3+y^3)+(21x+21y)\\=(x+y)(x^2-xy+y^2)+21(x+y)\\=(x+y)(x^2-xy+y^2+21)\)

#\(Toru\)

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

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

3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

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

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

3) \(x^2+7x+12=\left(x^2+3x\right)+\left(4x+12\right)=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

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

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

3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

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

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

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

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

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

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

1) \(x^2-4xy+4y^2+xz-2yz\)

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

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

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

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

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

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

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

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: \(x^2-6x-y^2+9\)

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

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

1. \(=3x\left(2x+5\right)\)

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

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

1, = 3x.(2x + 5)

2. = (3x)² - 1² = (3x - 1).(3x +1 )

3, =(x² + 6x + 9) - y² = (x + 3)² - y² =(x + y -3 ). (x - y +3)

3.\(x^2+6x-y^2+9=\left(x^2+6x+9\right)-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)

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: \(3x^2-11x+6\)

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

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

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