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29 tháng 8 2023

\(a,2x^2+3x-2xy-3y\)

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

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

\(b,x^3-4x^2+4x\)

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

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

#Urushi

29 tháng 8 2023

a) \(2x^2+3x-2xy-3y\)

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

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

b) \(x^3-4x^2+4x\)

\(\text{=}x\left(x^2-4x+4\right)\)

\(\text{=}x\left(x-2\right)^2\)

14 tháng 10 2020

21, \(x^3-4x^2+4x=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\)

22, \(15x^2y+20xy^2-25xy=5xy\left(3x+4y-5\right)\)

23, \(4x^2+8xy-3x-6y=4x\left(x+2y\right)-3\left(x+2y\right)=\left(4x-3\right)\left(x+2y\right)\)

24, \(x^3-6x^2+9x=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)

Tương tự :)) 

14 tháng 10 2020

21.\(x^3-4x^2+4x\)

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

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

22,\(15x^2y+20xy^2-25xy\)

\(=5xy\left(3x+4y-5\right)\)

23,\(4x^2+8xy-3x-6y\)

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

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

24\(x^3-6x^2+9x\)

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

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

25,\(x^2-xy+x-y\)

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

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

26.\(xy-2x-y^2+2y\)

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

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

27,\(x^2+x-xy-y\)

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

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

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

28,\(x^2+4x-y^2+4\)

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

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

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

29.\(x^2-2xy+y^2-4\)

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

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

14 tháng 10 2020

1) 2x2 - 4x = 2x( x - 2 )

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

3) x2 - 3x = x( x - 3 )

4) 4x2 - 6x = 2x( x - 3 )

5) x3 - 4x = x( x2 - 4 ) = x( x - 2 )( x + 2 )

14 tháng 10 2020

1) \(2x^2-4x=2x\left(x-2\right)\)

2) \(3x-6y=3\left(x-2y\right)\)

3) \(x^2-3x=x\left(x-3\right)\)

4) \(4x^2-6x=2x\left(2x-3\right)\)

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

1 tháng 9 2023

a) \(3x^2-3xy-5x+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

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

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

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

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

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

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

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

29 tháng 9 2019

cái này trong sách í

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

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

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

Đa thức này ko phân tích được nha bạn

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

\(=\left(x+1\right)\cdot\left(x+1\right)-\left(x+1\right)\cdot3\)

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

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

3: \(2x\cdot\left(x-2\right)-\left(x-2\right)^2\)

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

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

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

4: \(3x\left(x-1\right)^2-\left(1-x\right)^3\)

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

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

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

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

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

\(=\left(x+2\right)\cdot3x-\left(x+2\right)\cdot\left(5x+10\right)\)

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

\(=\left(-2x-10\right)\left(x+2\right)\)

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

6: \(4x\left(x-y\right)+3\left(y-x\right)^2\)

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

\(=\left(x-y\right)\cdot4x+\left(x-y\right)\left(3x-3y\right)\)

\(=\left(x-y\right)\cdot\left(4x+3x-3y\right)\)

\(=\left(x-y\right)\left(7x-3y\right)\)

4 tháng 12 2023

Cảm ơn nhiều

9 tháng 10 2021

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

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

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

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

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

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

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

d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)

a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)

\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)

b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)

c) Ta có: \(-4x^2+12xy-9y^2+25\)

\(=-\left(4x^2-12xy+9y^2-25\right)\)

\(=-\left[\left(2x-3y\right)^2-25\right]\)

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

d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)

\(=\left(x-y\right)^2-\left(2m-n\right)^2\)

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

13 tháng 9 2020

a) Ta có: \(4x\left(2y-z\right)+7y\left(z-2y\right)\)

        \(=4x\left(2y-z\right)-7y\left(2y-z\right)\)

        \(=\left(4x-7y\right)\left(2y-z\right)\)

13 tháng 9 2020

b) Ta có: \(2x\left(x+3\right)+\left(3+x\right)\)

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