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Vy trần
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Lấp La Lấp Lánh
10 tháng 10 2021 lúc 18:12

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

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

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

c) Đổi đề: \(a^2x+a^2y-7x-7y\)

\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)

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

e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

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

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

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

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

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Hoàng Anh Thắng
10 tháng 10 2021 lúc 18:14

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

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

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

e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

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

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

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

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ngọc hân
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ILoveMath
13 tháng 8 2021 lúc 10:09

a) x2y+xy+x+1= (x2y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)

b) x2-(a+b)x+ab=x2-ax-bx+ab=(x2-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)

c) ax2+ay-bx2-by=(ax2+ay)-(bx2+by)=a(x2+y)-b(x2+y)=(a-b)(x2+y)

d) ax-2x-a2+2a=(ax-2x)-(a2-2a)=x(a-2)-a(a-2)=(a-2)(x-a)

e) 2x2+4ax+x+2a=(2x2+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)

f) x3+ax2+x+a=(x3+ax2)+(x+a)=x2(x+a)+(x+a)=(x2+1)(x+a)

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Nguyễn Lê Phước Thịnh
13 tháng 8 2021 lúc 12:42

g: Ta có: \(x^4+2x^3-4x-4\)

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

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

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giúp mik với
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Nguyễn Lê Phước Thịnh
26 tháng 10 2021 lúc 21:24

a: \(=x\left(x-3\right)-4y\left(x-3\right)\)

=(x-3)(x-4y)

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

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

=2x(x+2)

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Nguyễn Hoàng Minh
26 tháng 10 2021 lúc 21:25

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

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hmone
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Tên ?
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Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:17

e) Ta có: \(x^4-2x^3+2x-1\)

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

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

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

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

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

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

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

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

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

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Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:11

a) Ta có: \(x^2-y^2-2x-2y\)

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

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

b) Ta có: \(x^2\left(x+2y\right)-x-2y\)

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

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

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Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:12

c) Ta có: \(x^3-4x^2-9x+36\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2-9\right)\)

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

d) Ta có: \(x^4+2x^3+2x-1\)

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

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

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Tên ?
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Trúc Giang
18 tháng 7 2021 lúc 16:36

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

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

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Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 22:59

a) Ta có: \(x^2-2x-4y^2-4y\)

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

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

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

b) Ta có: \(x^3+2x^2+2x+1\)

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

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

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

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Nguyễn Lê Phước Thịnh
18 tháng 7 2021 lúc 23:01

d) Ta có: \(a^6-a^4+2a^3+2a^2\)

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

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

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

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

c) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)

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

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

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Ẩn danh :)))
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HT.Phong (9A5)
14 tháng 9 2023 lúc 17:12

3) \(x^2\left(x+2y\right)-x-2y\)

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

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

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

4) \(x^3-4x^2-9x+36\)

\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)

\(=x^2\cdot\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2-9\right)\)

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

 

 

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\(x^2\left(x+2y\right)-x-2y\\ =x^2\left(x+2y\right)-\left(x+2y\right)\\ =\left(x^2-1\right)\left(x+2y\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ ---\\ x^3-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\ =\left(x^2-9\right)\left(x-4\right)\\ =\left(x-3\right)\left(x+3\right)\left(x-4\right)\)

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Lãng tử vô tình
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『Kuroba ム Tsuki Ryoo...
13 tháng 7 2023 lúc 17:42

`@` `\text {Ans}`

`\downarrow`

`a,`

`3x^2 + 6xy + 3y^2 - 3z`

`= 3*x^2 + 3*2xy + 3y^2 - 3z`

`= 3(x^2 + 2xy + y^2 - z)`

`b,`

`x^3 + x^2y - x^2z - xyz`

`= x(x + y)(x-z)`

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Mun SiNo
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Nguyễn Lê Phước Thịnh
11 tháng 10 2021 lúc 21:31

a: \(x^2-y^2-x-y\)

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

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

f: \(x^3-5x^2-5x+1\)

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

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

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