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1 tháng 2 2018

b)   \(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)

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

\(=x^6\left(x^3-1\right)-x^4\left(x^3-1\right)-x^2\left(x^3-1\right)+\left(x^3-1\right)\)

\(=\left(x^3-1\right)\left(x^6-x^4-x^2+1\right)\)

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

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

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

1 tháng 2 2018

       \(x^5+x+1\)

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

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

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

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

19 tháng 6 2021

 Nếu bn ko nhìn rõ thì ib vs mk nhé, mk sẽ gửi bản full cho bn.

19 tháng 6 2021

Hình nhéĐấy nhé

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

16 tháng 5 2021

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

19 tháng 7 2021

Trả lời:

Bài 4:

b, B =  ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 ) 

= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1 

= x8 - 1

Thay x = 2 vào biểu thức B, ta có:

28 - 1 = 255

c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 ) 

= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1

= x7 + 1

Thay x = 2 vào biểu thức C, ta có:

27 + 1 = 129

d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 ) 

= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x

= x

Thay x = - 5 vào biểu thức D, ta có:

D = - 5

Bài 5: 

a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )

= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4

= x4 - y4

Thay x = 2; y = - 1/2 vào biểu thức A, ta có:

A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16

b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 ) 

= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5 

= a5 + a4b - ab4 - b5

Thay a = 3; b = - 2 vào biểu thức B, ta có:

B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65

c, ( x2 - 2xy + 2y2 ) ( x+ y) + 2x3y - 3x2y+ 2xy3 

= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y+ 2xy3

= x4 + 2y4

Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:

( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16

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) . 

25 tháng 9 2017

1

x3-7x+6

=x3+0x2-7x +6

= x3-x2+x2-x-6x+6

=(x3-x2)+(x2-x)-(6x-6)

=x2(x-1)+x(x-1)-6(x-1)

=(x-1)(x2+x-6)

=(x-1)(x2+3x-2x-6)

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

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

25 tháng 9 2017

7) (x+2)(x+3)(x+4)(x+5)-24

=(x+2)(x+5) (x+3)(x+4)-24

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

=[x2+5x+2x+10][x2+4x+3x+12]-24

=[x2+7x+10][x2+7x+12]-24

đặt a=x2+7x+10

=>x2+7x+12=a+2

=a(a+2)-24

=a2+2a-24

=a2+6a-4a-24

=(a2+6a)-(4a+24)

=a(a+6)-4(a+6)

=(a+6)(a-4)

thay a= x2+7x+10 vào ta được

(x2+7x+10+6)(x2+7x+10-4)

=(x2+7x+16)(x2+7x+6)

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})`