K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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})`

6 tháng 7 2017

1) Ta có : 2x+ 3x - 5

= 2x2 - 2x + 5x - 5

= 2x(x - 1) + 5(x - 1)

= (x - 1) (2x + 5) 

3) x2 + x - 6

= x2 + 2x - 3x - 6

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

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

= (x - 3)(x + 2) 

23 tháng 7 2023

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

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

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

25 tháng 9 2017

Ta có : x3 - 7x + 6 

= x3 - x - 6x + 6 

= x(x2 - 1) - 6(x - 1)

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

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

= (x - 1) (x2 + x - 6) . 

CÁC Ý SAU TƯƠNG TỰ

19 tháng 2 2018

   x3 - 7x + 6 

= x3 - x - 6x + 6 

= x(x2 - 1) - 6(x - 1)

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

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

= (x - 1) (x2 + x - 6) . 

21 tháng 9 2019

a/\(\left(x^2-x\right)^2+4\left(x^2-x\right)-12.\)

cho \(\left(x^2-x\right)=a\)

\(\Rightarrow a^2+4a-12\)

\(=a^2+6a-2a-12\)

\(=\left(a^2+6a\right)-\left(2a+12\right)\)

\(=a\left(a+6\right)-2\left(a+6\right)\)

\(=\left(a+6\right)\left(a-2\right)\)

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

b/ \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Gọi \(x^2+5x+5=a\)

\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=\left(a-1\right)\left(a+1\right)-24\)

                                                                                 \(=a^2-1-24\)

                                                                                \(=a^2-25\)

                                                                                \(=\left(a-5\right)\left(a+5\right)\)

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

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

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

21 tháng 9 2021

\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)

21 tháng 9 2021

quá đỉnh luôn hehe