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22 tháng 5 2018

22 tháng 7 2023

a) \(x^2-xy+x-y\)

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

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

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

b) \(x^2+2xy-4x-8y\)

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

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

c) \(x^3-x^2-x+1\)

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

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

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

1A. Phân tích các đa thức sau thành nhân tử:a) x3+2x;                                           b) 3x - 6y;c) 5(x + 3y)- 15x(x + 3y);               d) 3(x-y)- 5x(y-x).1B. Phân tích các đa thức sau thành nhân tử:a) 4x2 - 6x;                                         b) x3y - 2x2y2 + 5xy; c) 2x2(x +1) + 4x(x +1);               d) 2 x(y - 1) - 2 y(1 - y). 5                  52A. Phân tích các đa thức sau thành nhân tử: a) 2(x -1)3 - 5(x -1)2 - (x - 1);b) x(y - x)3 - y(x - y)2 + xy(x -...
Đọc tiếp

1A. Phân tích các đa thức sau thành nhân tử:

a) x3+2x;                                           b) 3x - 6y;

c) 5(x + 3y)- 15x(x + 3y);               d) 3(x-y)- 5x(y-x).

1B. Phân tích các đa thức sau thành nhân tử:

a) 4x2 - 6x;                                         b) x3y - 2x2y2 + 5xy;

 

c) 2x2(x +1) + 4x(x +1);               d) 2 x(y - 1) - 2

 

y(1 - y).

 

5                  5

2A. Phân tích các đa thức sau thành nhân tử: a) 2(x -1)3 - 5(x -1)2 - (x - 1);

b) x(y - x)3 - y(x - y)2 + xy(x - y);

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

d) x(x + y)2 - y(x + y)2 + y2 (x - y).

2B. Phân tích đa thức thành nhân tử: a) 4(2-x)2 + xy - 2y;

b) x(x- y)3 - y(y - x)2 - y2(x - y);

c) x2y-xy2 - 3x + 3y;

d) x(x + y)2 - y(x + y) 2 + xy - x 2 .

3

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

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

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

7 tháng 10 2021

1A. a. x(x2+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x2-2xy+5)

c. 2x(x+1)(x+2)

d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)

 

19 tháng 8 2021

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt \(t=x^2+5x+4\)

(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)

Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)

a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)

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

\(=\left(x+15\right)\left(3x-4\right)\)

`a, a^3 - a^2b + a - b`

`= a^2(a-b) + (a-b)`

`= (a^2+1)(a-b)`

`b, x^2 - y^2 + 2y - 1`

`= x^2 - (y-1)^2`

`= (x-y+1)(x+y-1)`

a: \(=\dfrac{2}{5}\left(xy-x-y^2+1\right)\)

\(=\dfrac{2}{5}\left[x\left(y-1\right)-\left(y-1\right)\left(y+1\right)\right]\)

\(=\dfrac{2}{5}\left(y-1\right)\left(x-y-1\right)\)

b: \(=x\left(x^2+2xy+y^2-9\right)\)

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

2 tháng 4 2022

a) =25(xy−x−y2+1)

=25[x(y−1)−(y−1)(y+1)]

=25(y−1)(x−y−1)

b) =x(x2+2xy+y2−9)

23 tháng 7 2023

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

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

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

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

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

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

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

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

30 tháng 8 2021

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, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`

`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`

`c, xy^2 + x^2y + 1/4y^3`

`= y(xy + x^2 + 1/4y^2)`

`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`

`= (x+1+y)(x+1-y)`

13 tháng 11 2021

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

b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)

c) đặt \(x^2+x+1=t\)

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

10 tháng 9 2023

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