\(a.3x^2-3y^2-2\left(x-y\right)^2\\ =3\left(x^2-y^2\right)-2\left(x-y\right)^2\\ =3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\\ =\left(x-y\right)\left[3\left(x+y\right)-2.\left(x-y\right)\right]=\left(x-y\right)\left(x+5y\right)\\ b.x^2-y^2-2x-2y\\ =\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x-y-2\right)\\ c.\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\\ =\left(x-1\right)\left(2x+1\right)\left[1+3\left(x+2\right)\right]\\ =\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\\ d.\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\\ =\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\\ =\left(x-5\right)\left[\left(x-5\right)+\left(x+5\right)+\left(2x+1\right)\right]\\ =\left(x-5\right)\left(4x+1\right)\)

a) 3x²-3y²-2(x-y)²

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

a: \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

\(=\left(x-y\right)\left(x+5y\right)\)

b: Ta có: \(x^2-y^2-2x-2y\)

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

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

Bài 1 yêu cầu gì em?

Bài 2:

\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)

Bài 1:

a) \(3xy+6y\)

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

b) \(3x^2+9x\)

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

c) \(6x-9y^2\)

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

d) \(10xy^2-6x^2y\)

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

Bài 2:

a) \(x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(x+5\right)\)

b) \(3x\left(x+1\right)+3\left(x+1\right)\)

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

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

\(=3\left(x+1\right)^2\) 

c) \(x\left(x-3\right)+xy\left(x-3\right)\)

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

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

d) \(2x\left(x-2\right)-6\left(x-2\right)\)

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

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

Bài `1`

`a,3xy +6y`

`= 3y(x+2)`

`b,3x^2+9x`

`= 3x(x+3)`

`c,6x-9y^2`

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

`d,10xy^2-6x^2y`

`= 2xy(5y-3x)`

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

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

\(=\left(x-3\right)\left(x-4\right)\)

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

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

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

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

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

\(=\left(x-5\right)(x-5+5+x+2x+1)\)

\(=\left(x-5\right)\left(4x+1\right)\)

Còn lại bạn tự làm nhá

x^10 + x^5 + 1 
= x^10 + x^9 - x^9 + x^8 - x^8 + x^7 - x^7 + x^6 - x^6 + x^5 + x^5 - x^5 + x^4 - x^4 + x^3 - x^3 + x^2 - x^2 + x - x + 1 
= (x^10 + x^9 + x^8) - (x^9 + x^8 + x^7) + (x^7 + x^6 + x^5) - (x^6 + x^5 + x^4) + (x^5 + x^4 + x^3) - (x^3 + x^2 + x) + (x^2 + x + 1) 
= x^8 (x^2 + x + 1) - x^7 (x^2 + x + 1) + x^5 (x^2 + x + 1) - x^4 (x^2 + x + 1) + x^3 (x^2 + x + 1) - x (x^2 + x + 1) + (x^2 + x + 1) 
= (x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) 
----------------------- 
Phương pháp: 
Khi gặp bài toán phân tích thành nhân tử dạng x^(3m + 1) + x^(3n + 2) + 1 em thêm bớt các hạng tử từ bậc cao nhất trừ đi 1 đến x (bậc nhất) sao cho tổng số các hạng tử trong đa thức mới là một bội của 3. Sau đó nhóm ba hạng tử một sao cho trong mỗi nhóm có x² + x + 1 
Dạng này khi phân tích luôn có kết quả là: (x² + x + 1).Q(x)

x^7 + x^2 + 1 = x^7 + x^6 - x^6 + x^5 - x^5 + x^4 - x^4 +x^3 - x^3 +2x^2 - x^2 +x - x +1 
=(x^7 + x^6 + x^5) - (x^6 +x^5 +x^4) + (x^4 + x^3 +x^2) - (x^3 +x^2 + x) + (x^2 + x +1) 
=x^5(x^2 + x + 1) - x^4(x^2 + x + 1) +x^2(x^2 + x + 1) - x(x^2 + x + 1) + (x^2 + x + 1) 
=(x^2 + x + 1)(x^5 - x^4 +x^2 -x +1)

Các phần về đa thức bạn có thể vào đây giải trực tuyến luôn bạn nhé.

http://ungdungtoan.com/

Bài làm:

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

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

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

b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)

c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Làm mẫu cho 1 vd:

a, (x+1)(x+2)(x+3)(x+4)+1

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

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(1)

Đặt \(y=x^2+5x+5\)

Khi đó ::

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

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

Thay vào ta được: \(\left(x^2+5x+5\right)^2\)

a) (x+1)(x+2)(x+3)(x+4)+1=[(x+1)(x+4)].[(x+2)(x+3)]+1=(x²+5x+4)(x²+5x+6)+1

đặt t=x²+5x+5 ta có đa thức (t-1)(t+1)+1=t²-1+1=t². mà t=x²+5x+5

=> (x+1)(x+2)(x+3)(x+4)+1=(x²+5x+5)²

b) (x+1)(x+2)(x+3)(x+4)-24. theo kết quả câu (a) ta được (x+1)(x+2)(x+3)(x+4)=(x²+5x+4)(x²+5x+6)

đặt t=x²+5x+5 ta có đa thức (t-1)(t+1)-24=t²-1-24=t²-25=(t-5)(t+5)

mà t=x²+5x+5 => (t-5)(t+5)=(x²+5x)(x²+5x+10)

c) (x+1)(x+3)(x+5)(x+7)+15=[(x+1)(x+7)].[(x+3)(x+5)]+15=(x²+8x+7)(x²+8x+15)+15

đặt x²+8x+11=t ta có đa thức (t-4)(t+4)+15=t²-16+15=t²-1=(t-1)(t+1)

mà t=x²+8x+11 => (t-1)(t+1)=(x²+8x-10)(x²+8x+12)

d) (x+2)(x+3)(x+4)(x+5)-24=[(x+2)(x+5)][(x+3)(x+4)]-24=(x²+7x+12)(x²+7x+10)-24

đặt t=x²+7x+11 ta có đa thức (t-1)(t+1)-24=t²-1-24=t²-25=(t+5)(t-5)

mà t=x²+7x+11 => (t-5)(t+5)=(x²+7x+6)(x²+7x+16)

a: Ta có: \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)

\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)

\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)

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

b: Ta có: \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)\)

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

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

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

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

a) \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)

b) \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)=\left(x-y\right)\left[-x\left(x-y\right)^2-y\left(x-y\right)+xy\right]=\left(x-y\right)\left(-x^3+2x^2y-xy^2-xy+y^2+xy\right)=\left(x-y\right)\left(-x^3+y^2+2x^2y-xy^2\right)\)

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

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

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

Lời giải:

a.

$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$

b.

$=\frac{1}{3}x(y+3xz+3z)$

c.

$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$

d.

$=\frac{1}{5}x^2(2+25x+5y)$

\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

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

\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)