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Dương
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ILoveMath
31 tháng 7 2021 lúc 16:33

a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)

b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)

c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

Nguyễn Lê Phước Thịnh
31 tháng 7 2021 lúc 23:10

a) Ta có: \(x^3+4x-5\)

\(=x^3-x+5x-5\)

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

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

b) Ta có: \(x^3-3x^2+4\)

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

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

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

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

c) Ta có: \(x^3+2x^2+3x+2\)

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

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

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

d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)

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

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

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

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

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

e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)

\(=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8\)

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

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

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

f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)

\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)

\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)

\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)

\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)

Cíu iem
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ILoveMath
10 tháng 10 2021 lúc 16:59

a) \(xy^2-25x=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\)

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

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

Nguyễn Hoàng Minh
10 tháng 10 2021 lúc 16:59

\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=x\left(x-y\right)+2\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)

nthv_.
10 tháng 10 2021 lúc 17:01

\(x\left(y^2-25\right)=y\left(y-5\right)\left(y+5\right)\)

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

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

ThanhNghiem
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๖ۣۜHả๖ۣۜI
4 tháng 8 2023 lúc 14:25

\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)

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

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

\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)

 

 

 

Nguyễn Lê Phước Thịnh
4 tháng 8 2023 lúc 14:15

b: =xy-x-y+1

=x(y-1)-(y-1)

=(x-1)(y-1)

c: =(x-2y)^2-4y

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

d: =16-(x^2-2xy+y^2)

=16-(x-y)^2

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

Thư Vũ
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Nguyễn Việt Lâm
14 tháng 11 2021 lúc 14:22

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

\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)

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

\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)

giúp mik với
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Nguyễn Hoàng Minh
26 tháng 10 2021 lúc 21:12

\(a,=y\left(y-2\right)\\ b,=3x\left(x^2-2x+1\right)=3x\left(x-1\right)^2\\ c,=\left(y-1\right)\left(27x^2+9x^3\right)=9x^2\left(x+3\right)\left(y-1\right)\\ d,=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\\ e,=x\left(x^2+6x+9\right)=x\left(x+3\right)^2\\ f,=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\\ g,=\left(2-x\right)\left(x+1\right)\\ h,=\left(x-1\right)\left(3x-6\right)=3\left(x-1\right)\left(x-2\right)\)

Nguyễn Lê Phước Thịnh
26 tháng 10 2021 lúc 21:14

a: =y(y-2)

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

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

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38. Nguyễn Diệp Minh Thy
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ngọc hân
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Nguyễn Hoàng Minh
19 tháng 9 2021 lúc 15:58

\(a,=\left(2x-5\right)\left(x+1\right)\\ b,=\left(x-10\right)\left(x+1\right)\\ c,=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

Minh Trí Nguyễn
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Nguyễn Việt Lâm
28 tháng 12 2021 lúc 19:44

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

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

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

Bơ Ngố
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ILoveMath
2 tháng 1 2022 lúc 21:28

\(a,x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ v,x^3-2x^2-x+2=\left(x^3-2x^2\right)-\left(x-2\right)=x^2\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(x^2-1\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\\ c,25-16x^2=\left(5-4x\right)\left(5+4x\right)\)

Nguyễn Hoàng Minh
2 tháng 1 2022 lúc 21:28

\(a,=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ b,=x^2\left(x-2\right)-\left(x-2\right)\\ =\left(x-2\right)\left(x^2-1\right)=\left(x-1\right)\left(x-2\right)\left(x+2\right)\\ c,=\left(5-4x\right)\left(5+4x\right)\)

Nguyễn Lê Phước Thịnh
2 tháng 1 2022 lúc 21:29

c: =(5-4x)(5+4x)