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12 tháng 9 2021

\(x^2-x-2001.2002=x^2-x-4006002\)

12 tháng 9 2021

nhân đa thức thành nhân tử

12 tháng 9 2021

\(x^2-x-2001.2002=x^2-2002x+2001x-2001.2002=x\left(x-2002\right)+2001\left(x-2002\right)=\left(x+2001\right)\left(x-2002\right)\)

12 tháng 9 2021

bài dễ mà bạn mình ra cho mấy bạn kiếm SP sau giờ mới làm non thế

24 tháng 8 2019

\(x^2-x-2001.2002\)

\(x^2+2001x-2002x-2001.2002\)

\(x\left(x+2001\right)-2002\left(x+2001\right)\)

\(\left(x+2001\right)\left(x-2002\right)\)

a) Ta có: \(x^3+x^2+4\)

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

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

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

b) Ta có: \(9x^2+12x-5\)

\(=9x^2+15x-3x-5\)

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

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

c) Ta có: \(x^4+1997x^2+1996x+1997\)

\(=x^4+x^2+1+1996x^2+1996x+1996\)

\(=\left(x^4+2x^2+1-x^2\right)+1996\left(x^2+x+1\right)\)

\(=\left[\left(x^2+1\right)^2-x^2\right]+1996\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

d) Ta có: \(x^2-x-2001\cdot2002\)

\(=x^2-2002x+2001x-2001\cdot2002\)

\(=x\left(x-2002\right)+2001\left(x-2002\right)\)

\(=\left(x-2002\right)\left(x+2001\right)\)

6 tháng 7 2021

Cảm ơn bạn nhiều ❤️❤️❤️

16 tháng 5 2019

a) \(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)

\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2y+x^2z+xyz-xyz\)

\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2\left(y+z\right)\)

\(=\left(y+z\right)\left(xy+yz+zx+x^2\right)\)

\(=\left(y+z\right)\left[y\left(x+z\right)+x\left(z+x\right)\right]\)

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

b) \(\left(x^2+y^2+5\right)^2-4x^2y^2-16xy-16\)

\(=\left(x^2+y^2+5\right)^2-\left(4x^2y^2+16xy+16\right)\)

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

\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)

\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)

16 tháng 5 2019

c)sai đề. 

đặt \(x^2+x+1=t\)

\(\Rightarrow\left(x^2+x+1\right)^2+\left(x^2+x+2\right)-12\)

\(=t^2+t+1-12\)

.........................................

mình sửa đề không biết có đúng hay không nên mình chỉ nêu hướng làm thôi. bạn thông cảm.

d) \(x^2-x-2001.2002\)

\(=x\left(x+2001\right)-2002\left(x+2001\right)\)

\(=\left(x-2002\right)\left(x+2001\right)\)

24 tháng 6 2017

a) Ta có : x2 - 4x + 3

= x2 - x - 3x + 3

= x(x - 1) - (3x - 3) 

= x(x - 1) - 3(x - 1)

= (x - 1) (x - 3) 

24 tháng 6 2017

a) \(x^2-4x+3\)

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

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

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

b) \(x^2+5x+4\)

\(=x^2+x+4x+4\)

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

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

c) \(x^2-x-6\)

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

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

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

d) \(x^4+1997x^2+1996x+1997\)

\(=x^4+x^2+1996x^2+1996x+1996+1\)

\(=\left(x^4+x^2+1\right)+\left(1996x^2+1996x+1996\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

e) \(x^2-2001\cdot2002\)( hình như sai sai)

20 tháng 12 2020

1, \(45+x^3-5x^2-9x=9\left(5-x\right)+x^2\left(x-5\right)\)

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

3, \(x^4-5x^2+4\)

Đặt \(x^2=t\left(t\ge0\right)\)ta có : 

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

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

29 tháng 3 2022

`Answer:`

1. `45+x^3-5x^2-9x`

`=x^3+3x^2-8x^2-24x+15x+45x`

`=x^2 .(x+3)-8x.(x+3)+15.(x+3)`

`=(x+3).(x^2-8x+15)`

`=(x+3).(x^2-5x-3x+15)`

`=(x-3).(x-5).(x-3)`

2. `x^4-2x^3-2x^2-2x-3`

`=x^4+x^3-3x^3+x^2+x-3x-3`

`=x^3 .(x+1)-3x^2 .(x+1)+x.(x+1)-3.(x+1)`

`=(x+1).(x^3-3x^2+x-3)`

`=(x+1).[x^3 .(x-3).(x-3)]`

`=(x+1).(x-3).(x^2+1)`

3. `x^4-5x^2+4`

`=x^4-x^2-4x^2+4`

`=x^2 .(x^2-1)-4.(x^2-1)`

`=(x^2-1).(x^2-4)`

`=(x-1).(x+1).(x-2).(x+2)`

4. `x^4+64`

`=x^4+16x^2+64-16x^2`

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

`=(x^2+8-4x).(x^2+8+4x)`

5. `x^5+x^4+1`

`=x^5+x^4+x^3-x^3+1`

`=x^3 .(x^2+x+1)-(x^3-1)`

`=x^3 .(x^2+x+1)-(x-1).(x^2+x+1)`

`=(x^2+x+1).(x^3-x+1)`

6. `(x^2+2x).(x^2+2x+4)+3`

`=(x^2+2x)^2+4.(x^2+2x)+3`

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

`=(x^2+2x+1).(x^2+2x)+3.(x^2+2x+1)`

`=(x^2+2x+1).(x^2+2x+3)`

`=(x+1)^2 .(x^2+2x+3)`

7. `(x^3+4x+8)^2+3x.(x^2+4x+8)+2x^2`

`=x^6+8x^4+16x^3+16x^2+64x+64+3x^3+12x^2+24x+2x^2`

`=x^6+8x^4+19x^3+30x^2+88x+64`

8. `x^3 .(x^2-7)^2-36x`

`=x[x^2.(x^2-7)^2-36]`

`=x[(x^3-7x)^2-6^2]`

`=x.(x^3-7x-6).(x^3-7x+6)`

`=x.(x^3-6x-x-6).(x^3-x-6x+6)`

`=x.[x.(x^2-1)-6.(x+1)].[x.(x^2-1)-6.(x-1)]`

`=x.(x+1).[x.(x-1)-6].(x-1).[x.(x+1)-6]`

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

`=x.(x+1).(x-1).[x.(x-3)+2.(x-3)].[x.(x+3)-2.(x+3)]`

`=x.(x+1)(x-1).(x-2).(x+2).(x-3).(x+3)`

9. `x^5+x+1`

`=x^5-x^2+x^2+x+1`

`=x^2 .(x^3-1)+(x^2+x+1)`

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

`=(x^2+x+1).(x^3-x^2+1)`

10. `x^8+x^4+1`

`=[(x^4)^2+2x^4+1]-x^4`

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

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

`=[(x^4+2x^2+1)-x^2].(x^4-x^2+1)`

`=[(x^2+1)^2-x^2].(x^4-x^2+1)`

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

11. ` x^5-x^4-x^3-x^2-x-2`

`=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2`

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

`=(x-2).(x^4+x^3+x^2+x+1)`

12. `x^9-x^7-x^6-x^5+x^4+x^3+x^2-1`

`=(x^9-x^7)-(x^6-x^4)-(x^5-x^3)+(x^2-1)`

`=x^7 .(x^2-1)-x^4 .(x^2-1)-x^3 .(x^2-1)+(x^2-1)`

`=(x^2-1).(x^7-x^4-x^3+1)`

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

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

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

`=(x-1)^3 .(x+1)^2 .(x^2+x+1)(x^2+1)`

13. `(x^2-x)^2-12(x^2-x)+24`

`=[ (x^2-x)^2-2.6(x^2-x)+6^2]-12`

`=(x^2-x+6)^2-12`

`=(x^2-x+6-\sqrt{12})(x^2-x+6+\sqrt{12})`

AH
Akai Haruma
Giáo viên
11 tháng 8 2021

1.

$(x-2)(x-5)=(x-3)(x-4)$

$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)

Vậy pt vô nghiệm.

2.

$(x-7)(x+7)+x^2-2=2(x^2+5)$

$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$

$\Leftrightarrow -51=10$ (vô lý)

Vậy pt vô nghiệm.

AH
Akai Haruma
Giáo viên
11 tháng 8 2021

3.

$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$

$\Leftrightarrow 4x+10=-8$

$\Leftrightarrow 4x=-18$

$\Leftrightarrow x=-4,5$

4.

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

$\Leftrightarrow x^2+2x+1=x^2+x-6$

$\Leftrightarrow x=-7$