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giúp mik với
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Nguyễn Lê Phước Thịnh
26 tháng 10 2021 lúc 21:36

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

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

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

c: \(=x\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(x-10\right)\)

Lấp La Lấp Lánh
26 tháng 10 2021 lúc 21:39

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

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

c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)

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

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

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

g) \(=y\left(y^2-2xy+x^2-y\right)\)

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

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

02 Nguyễn Hoàng Chương
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☆Châuuu~~~(๑╹ω╹๑ )☆
18 tháng 1 2022 lúc 13:50

\(a,x\left(x+6\right)\\ b,\left(9x-1\right)\left(9x+1\right)\\ c,\left(x+y\right)-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ d,\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\)

Phương Nhi
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Lê Trang
27 tháng 12 2020 lúc 9:21

a) \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\left(2x-3y+4xy\right)\)

b) \(3x^2-5x-3xy+5y\)

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

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

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

c) \(5a^3-20a\)

\(=5a\left(a^2-4\right)\)

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

d) \(2x+2y+x^2+2xy+y^2\)

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

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

NT Quỳnh Anh
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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)

02 Nguyễn Hoàng Chương
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ILoveMath
10 tháng 12 2021 lúc 18:15

\(a,4x-20y=4\left(x-5y\right)\\ b,10x^2+10xy-x-y=10x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(10x-1\right)\\ c,x^2-2xy-z^2+y^2=\left(x^2-2xy+y^2\right)-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

Nguyễn Trần Nhật Khang
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Nguyễn Hoàng Minh
26 tháng 11 2021 lúc 14:09

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

Mineru
26 tháng 11 2021 lúc 14:10

a) 4x - 20y

= 4 ( x - 5y )

b) 5x^2 + 5xy - x - y

= 5x ( x + y ) - ( x - y )

= ( x + y ) ( 5x - 1 )

c) x^2 - 2xy - z^2 + y^2

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

= ( x - y )^2 - z^2

= ( x - y + z ) ( x - y - z )

Rhider
26 tháng 11 2021 lúc 14:10

a) \(4\left(x-5y\right)\)

b) \(5x^2+5xy-x-y\)

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

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

c) \(x^2-2xy-z^2+y^2\)

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

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

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

Cíu iem
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Rin Huỳnh
21 tháng 9 2021 lúc 11:33

a) 3(x-y) - (x-y)^2

 =(x-y)(3-x+y)

Rin Huỳnh
21 tháng 9 2021 lúc 11:34

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

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

Nguyễn Hoàng Minh
21 tháng 9 2021 lúc 11:35

\(a,=3\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(3-x+y\right)\\ b,=\left(x+y\right)^2-4x^2y^2=\left(x-2xy+y\right)\left(x+2xy+y\right)\\ c,=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)

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

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

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

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

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