K
Khách

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

18 tháng 8 2016

a,x2 - 13x + 36

= x- 4x - 9x + 36

= x(x - 4) - 9(x - 4)

= (x - 4)(x - 9)

b,x2 + 3x - 18

= x- 9 + 3x - 9

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

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

= (x - 3)(x + 6)

c,x2 - 5x - 24

= x2 + 3x - 8x - 24

= x(x + 3) - 8(x + 3)

= (x + 3)(x - 8)

d,3x2 - 16x + 5

= 3x2 - 15x - x + 5

= 3x(x - 5) - (x - 5)

= (x - 5)(3x - 1)

e, 8x2 + 30x + 7

= 8x2 + 2x + 28x + 7

= 2x(4x + 1) + 7(4x + 1)

= (4x + 1)(2x + 7)

g,2x2 - 5x - 12

= 2x2 - 8x + 3x - 12

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

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

i,x4 + 4

= x4 + 4x2 + 4 - 4x2

= (x2 + 2)2 - (2x)2

= (x2 + 2 - 2x)(x2 + 2 + 2x)

4 tháng 9 2018

Phân tích đa thức thành nhân tử bằng cách tách một số hạng tử thành nhiều số hạng khác

a) x^2+4x+3

b) 4x^2-4x-3

c)x^2-x-12

d) 4x^4-4x^2-8y^4

27 tháng 7 2019

somebody help me 

27 tháng 7 2019

\(1,2x^2-3x-2\) 

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

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

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

\(2,4x^2-7x-2\)

\(=4x^2-8x+x-2\) 

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

\(\left(4x+1\right)\left(x-2\right)\)

14 tháng 8 2021

a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x-6y-1\right)\)

b) \(=x\left(x^2-8x+16\right)\)

\(=x\left(x-4\right)^2\)

c) \(=2\left(x-y\right)^2-18\)

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

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

a: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: \(x^3-8x^2+16x\)

\(=x\left(x^2-8x+16\right)\)

\(=x\left(x-4\right)^2\)

c: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

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

d: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

a: Ta có: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: Ta có: \(16x-8x^2+x^3\)

\(=x\left(x^2-8x+16\right)\)

\(=x\left(x-4\right)^2\)

c: Ta có: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

\(=2\cdot\left[\left(x-y\right)^2-9\right]\)

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

d: Ta có: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

e: Ta có: \(x^4-x^2-30\)

\(=x^4-6x^2+5x^2-30\)

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

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

f: Ta có: \(x^2-xy-2y^2\)

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

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

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

g: Ta có: \(x^4-13x^2y^2+4y^4\)

\(=x^4-4x^2y^2+4y^4-9x^2y^2\)

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

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

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

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

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

h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)

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

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

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

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

15 tháng 10 2021

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

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

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

c:\(-6x^2+7x-2\)

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

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

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

15 tháng 10 2021

1.

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

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

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

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

2.

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

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

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
7 tháng 7 2021

a. $6x^2-11x=x(6x-11)$
b. $x^7+x^5+1=(x^7-x)+(x^5-x^2)+x+x^2+1$

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

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

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

AH
Akai Haruma
Giáo viên
7 tháng 7 2021

c.

$x^8+x^4+1=(x^4)^2+2.x^4+1-x^4$

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

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

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

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

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

d.

$x^3-5x+8-4=x^3-5x+4$

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

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

e.

$x^5+x^4+1=(x^5-x^2)+(x^4-x)+x^2+x+1$

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

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

$=(x^2+x+1)[(x-1)(x^2+x)+1]$

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

 

 

6 tháng 9 2020

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)

b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)

c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)

d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)

6 tháng 9 2020

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)

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

b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)

\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)

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

c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)

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

d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)

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

27 tháng 10 2021

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

4 tháng 9 2016

a) x2-xz-9y2+3yz

=(x2-9y2)-(xz-3yz)

=(x-3y)(x+3y)-z(x-3y)

=(x-3y)(x+3y-z)

b)x3-x2-5x+125

=x3-6x2+25x+5x2-30x+125

=x(x2-6x+25)+5(x2-6x+25)

=(x+5)(x2-6x+25)

c.x3+2x2-6x-27

=x3+5x2+9x-3x2-15x-27

=x(x2+5x+9)-3(x2+5x+9)

=(x-3)(x2+5x+9)

d. 12x3+4x2-27x-9

=12x3+4x2-27x-9

=4x2(3x+1)-9(3x+1)

=(4x2-9)(3x+1)

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

e.x4-25x2+20x-4

=x4+5x3-2x2-5x2-25x+10+2x2+10x-4

=x2(x2+5x-2)-5(x2+5x-2)+2(x2+5x-2)

=(x2-5x+2)(x2+5x-2)

f.x2(x2-6)-x2+9

=x4+x3-3x2-x3-x2+3x-3x2-3x+9

=x2(x2+x-3)-x(x2+x-3)-3(x2+x-3)

=(x2-x-3)(x2+x-3)

5 tháng 9 2016

mà bạn ơi sao câu b bạn tách ra nv ?

26 tháng 12 2021

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

n: \(=4x^4+36x^2+81-36x^2\)

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