Những câu hỏi liên quan
Linda Ryna Daring
Xem chi tiết
4rUnME
5 tháng 10 2015 lúc 20:54

a) 

5x-5y+ax-ay = 5(x-y) +a(x-y) = (x-y)(5+a)

b) a^3 -a^2x-ay+xy = a^2(a-x) -y(a-x) = (a-x)(a^2-y)

c) xy(x+y) +yz(y+z) +xz(x+z) +2xyz = x^2.y+xy^2 +y^2.z+xz^2 +x^2.z+xz^2 +2xyz

= (x^2.y+x^2.z)+(xy^2+xz^2+2xyz)+(y^2.z+yz^2) = x^2(y+z) +x.(y+z)^2 +yz(y+z)

=(y+z)(x^2+x+yz)

Bình luận (0)
Nguyễn Thị Kim Anh
Xem chi tiết
Trần Anh
28 tháng 7 2017 lúc 16:56

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

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

3 ) \(5x-5y+ax-ay=5.\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)

4 ) \(a^3-a^2x-ay+xy=a^2.\left(a-x\right)-y.\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)

5 ) \(xy.\left(x+y\right)+yz.\left(y+z\right)+xz.\left(x+z\right)+2xyz\)

\(=xy.\left(x+y\right)+y^2z+yz^2+x^2z+xz^2+xyz+xyz\)

\(=xy.\left(x+y\right)+\left(y^2z+xyz\right)+\left(yz^2+xz^2\right)+\left(x^2z+xyz\right)\)

\(=xy.\left(x+y\right)+yz.\left(x+y\right)+z^2.\left(x+y\right)+xz.\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+yz+z^2+xz\right)=\left(x+y\right)\left[\left(xy+xz\right)+\left(yz+z^2\right)\right]\)

\(=\left(x+y\right)\left[x.\left(y+z\right)+z.\left(y+z\right)\right]=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Bình luận (0)
Die Devil
Xem chi tiết
Hoàng Phúc
3 tháng 9 2016 lúc 20:27

c) xét giá trị riêng

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+xyz+xyz\)

\(=xy\left(x+y\right)+y^2z+yz^2+x^2z+xz^2+xyz+xyz\)

\(=xy\left(x+y\right)+y^2z+xyz+yz^2+xz^2+x^2z+xyz\)

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

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

\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Bình luận (0)
Vũ Quang Vinh
3 tháng 9 2016 lúc 20:19

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

Bình luận (0)
Dương Tuệ Nhiên
3 tháng 9 2016 lúc 20:27

b) x- ax2 - xy + ay
=x3 -xy - ax+ay
=x(x2-y) - a(x2-y)
=(x-a)(x2-y)

Bình luận (0)
Trần Thu Phương
Xem chi tiết
Trần Thùy Dương
10 tháng 10 2018 lúc 22:53

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+xz\left(x+z\right)\)

\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)

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

\(=\left(x+z\right)\left(x^2+y^2+yz+xz\right)\)

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

Bình luận (0)
Nguyễn Xuân Anh
10 tháng 10 2018 lúc 22:56

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz.\)

\(=x^2y+xy^2+x^2z+xz^2+2xyz+yz\left(y+z\right)\)

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

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

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

\(=\left(y+z\right)\left[x\left(x+z\right)+y\left(x+z\right)\right]=\left(y+z\right)\left(x+y\right)\left(x+z\right)\)

Bình luận (0)
Kim Ngọc Huyền
Xem chi tiết
Trangeki
Xem chi tiết
Darlingg🥝
15 tháng 11 2019 lúc 21:34

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)2xyz\)

\(=\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+xz\left(x+z\right)\)

\(=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)

\(=\left(xy+yz\right)\left(x+y+z\right)+xz\left(x+z\right)\)

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

\(=\left(x+z\right)\left[y\left(x+y+z\right)+xz\right]=\left(x+z\right)\left(xy+y^2+yz+xz\right)\)

\(=\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]\)

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

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

Phức tạp. Cs cách nào nhanh kkk?

Bình luận (0)
 Khách vãng lai đã xóa
Cold Wind
Xem chi tiết
Kiệt ღ ๖ۣۜLý๖ۣۜ
27 tháng 6 2016 lúc 10:46

xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z2)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)

Bình luận (0)
Nguyễn Thị Quỳnh
Xem chi tiết
Cold Wind
Xem chi tiết
Bùi Thị Vân
27 tháng 6 2016 lúc 10:56

\(xy.\left(x+y\right)+yz.\left(y+z\right)+xz.\left(x+z\right)+2xyz\)
\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz\)

\(\Leftrightarrow xy\left(x+y\right)+xyz+yz\left(y+z\right)+xyz+xz\left(z+x\right)\)

\(\Leftrightarrow xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)\)

\(\Leftrightarrow y\left(x+y+z\right)\left(x+z\right)+xz\left(x+z\right)\)
\(\Leftrightarrow\left(x+z\right)\left(y\left(z+x\right)+zx\right)\)

\(\Leftrightarrow\left(x+z\right)\left(y+z\right)\left(x+y\right)\)

Bình luận (0)
Võ Đông Anh Tuấn
27 tháng 6 2016 lúc 10:43

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=xy.x+xy.y+yz.y+yz.z+xz.x+xz.z+2xyz\)

\(=x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz\)

Bình luận (0)