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Nguyễn Hữu Nguyên
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29 tháng 7 2021 lúc 10:13

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Nguyễn Lê Phước Thịnh
29 tháng 7 2021 lúc 13:43

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

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

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

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

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

Nguyễn Hữu Nguyên
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Nguyễn Huy Tú
6 tháng 8 2021 lúc 10:25

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

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

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

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

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

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

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

Nguyễn Lê Phước Thịnh
6 tháng 8 2021 lúc 11:20

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)

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

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

Minh Dũng Nguyễn
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Nguyễn Việt Lâm
22 tháng 10 2021 lúc 14:47

\(5x^2+10xy=5x\left(x+2y\right)\)

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

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

\(x^2-7x+6=x^2-x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

Phạm Như Hiếu
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Minh Hiếu
13 tháng 10 2021 lúc 16:11

a) \(2x^2+3x-27\)

\(=2x^2+9x-6x-27\)

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

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

b) sửa đề thành \(x^2+7x+6\)

\(x^2+7x+6\)

\(=x^2+x+6x+6\)

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

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

 

Phạm Như Hiếu
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Nguyễn Hoàng Minh
17 tháng 10 2021 lúc 8:10

\(a,=2x^2-6x+9x-27=\left(x-3\right)\left(2x+9\right)\\ b,=x^2-7x+\dfrac{49}{4}-\dfrac{73}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{73}{4}=\left(x-\dfrac{7}{2}-\dfrac{\sqrt{73}}{2}\right)\left(x-\dfrac{7}{2}+\dfrac{\sqrt{73}}{2}\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ d,=x^2-2x-8x+16=\left(x-2\right)\left(x-8\right)\\ e,=x^2-3x-5x+15=\left(x-3\right)\left(x-5\right)\\ g,=x^2+2x+4x+8=\left(x+2\right)\left(x+4\right)\)

Thuylinh
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HT.Phong (9A5)
22 tháng 7 2023 lúc 13:35

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

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

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

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

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

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

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

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

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

\(=\left(x-2\right)\left[2x-\left(x-2\right)\right]\)

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

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

Trần Mai
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Nguyễn Hoàng Minh
18 tháng 12 2021 lúc 21:48

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

Nguyễn Lê Phước Thịnh
18 tháng 12 2021 lúc 21:50

a: =(x-1)2

ngọc hân
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loann nguyễn
13 tháng 8 2021 lúc 11:21

\(a.3x^2-3y^2-2\left(x-y\right)^2\\ =3\left(x^2-y^2\right)-2\left(x-y\right)^2\\ =3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\\ =\left(x-y\right)\left[3\left(x+y\right)-2.\left(x-y\right)\right]=\left(x-y\right)\left(x+5y\right)\\ b.x^2-y^2-2x-2y\\ =\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x-y-2\right)\\ c.\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\\ =\left(x-1\right)\left(2x+1\right)\left[1+3\left(x+2\right)\right]\\ =\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\\ d.\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\\ =\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\\ =\left(x-5\right)\left[\left(x-5\right)+\left(x+5\right)+\left(2x+1\right)\right]\\ =\left(x-5\right)\left(4x+1\right)\)

 

linh phạm
13 tháng 8 2021 lúc 11:12

a) 3x2-3y2-2(x-y)2

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

Nguyễn Lê Phước Thịnh
13 tháng 8 2021 lúc 11:47

a: \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

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

b: Ta có: \(x^2-y^2-2x-2y\)

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

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

Trần Thái Sơn
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Nguyễn Hoàng Minh
19 tháng 10 2021 lúc 19:42

\(a,=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

ThyXingGái
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Nguyễn Hoàng Minh
15 tháng 12 2021 lúc 16:25

\(a,=x\left(x-2\right)^2\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\\ c,=x^2\left(2x-1\right)-4\left(2x-1\right)=\left(x-2\right)\left(x+2\right)\left(2x-1\right)\\ d,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ e,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x\left[\left(x-2\right)^2-y^2\right]=x\left(x-y-2\right)\left(x+y-2\right)\\ g,=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\\ h,=x^3-x-2x+2=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x-2\right)=\left(x-1\right)^2\left(x+2\right)\\ i,=3x^2+3x-10x-10=\left(x+1\right)\left(3x-10\right)\)