H
3
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
LN
2
MT
22 tháng 7 2015
A ) xy(z+y)+yz(y+z)+zx(z+x)
=y.[x(z+y)+z(y+z)]+zx(z+x)
=y.(xz+xy+zy+z2)+zx(z+x)
=y.(xz+z2+xy+zy)+zx(z+x)
=y.[z.(z+x)+y.(z+x)]+zx(z+x)
=y.(z+x)(z+y)+zx(z+x)
=(z+x)[y(z+y)+zx]
=(z+x)(yz+y2+zx)
B )xy(x+y)-yz(y+z)-zx(z-x)
=y.[x(x+y)-z(y+z)]-zx(z-x)
=y.(x2+xy-zy-z2)-zx(z-x)
=y.(x2-z2+xy-zy)-zx(z-x)
=y.[(x+z)(x-z)+y.(x-z)]-zx(z-x)
=y.(x-z)(x+z+y)+zx(x-z)
=(x-z)[y(x+z+y)+zx]
=(x-z)(yx+yz+y2+zx)
=(x-z)(yx+zx+yz+y2)
=(x-z)[x.(y+z)+y.(y+z)]
=(x-z)(y+z)(x+y)
LT
30 tháng 6 2021
b. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)
10 tháng 7 2017
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) Vô câu hỏi tương tự
26 tháng 7 2017
a) xy(x + y) + yz(z + y) + zx(z + x) + 3xyz
= [xy(x + y) + xyz] + [yz(z + y) + xyz] + [zx(z + x) + xyz]
= xy(x + y + z) + yz(x + y + z) + zx(x + y + z)
= (xy + yz + zx)(x + y + z)
b) tương tự
HD
1
MT
19 tháng 7 2015
xy(x+y)-yz(y+z)-zx(z-x)
=y.[x.(x+y)-z.(y+z)]-zx.(z-x)
=y.(x2+xy-zy-z2)-zx.(z-x)
=y.[(x-z)(x+z)-y.(z-x)]-zx.(z-x)
=y.[-(z-x)(x+z)-y.(z-x)]-zx.(z-x)
=y.(z-x)(-x-z-y)-zx.(z-x)
=(z-x)(-xy-zy-y2-zx)
=(z-x)[-x.(y+z)-y.(y+z)]
=(z-x)(y+z)(-x-y)
=-(z-x)(y+z)(x+y)
KY
16 tháng 9 2019
\(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2\)
\(=xy^2-xz^2+yz^2-x^2y+zx^2-zy^2-xyz+xyz\)
\(=\left(yz^2-xz^2-xyz+x^2z\right)-\left(zy^2-xyz-xy^2+x^2y\right)\)
\(=z\left(yz-xz-xy+x^2\right)-y\left(zy-xz-xy+x^2\right)\)
\(=\left(z-y\right)\left(yz-xz-xy+x^2\right)\)
\(=\left(z-y\right)\left[y\left(z-x\right)-x\left(z-x\right)\right]\)
\(=\left(z-y\right)\left(y-x\right)\left(z-x\right)\)
MC
0
TU
1
PA
29 tháng 10 2016
\(xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)\)
\(=xy\left(x+y\right)+z\left[y\left(y-z\right)-x\left(x+z\right)\right]\)
\(=xy\left(x+y\right)+z\left(y^2-yz-x^2-xz\right)\)
\(=xy\left(x+y\right)+z\left[\left(y^2-x^2\right)-\left(yz+xz\right)\right]\)
\(=xy\left(x+y\right)+z\left[\left(y-x\right)\left(y+x\right)-z\left(x+y\right)\right]\)
\(=xy\left(x+y\right)+z\left(x+y\right)\left(y-z-x\right)\)
\(=\left(x+y\right)\left[xy+z\left(y-x-z\right)\right]\)
\(=\left(x+y\right)\left(xy-yz-xz-z^2\right)\)
TU
1
29 tháng 10 2016
xy(x+y) + yz(y-z) - zx(z+x)
=y[x(x+y)+z(y-z)]-zx(z+x)
=y(x2+xy+yz-z2)-zx(z+x)
=y(x2-z2+xy+yz)-zx(z+x)
=y[(x-z)(x+z)+y(x+z)]-zx(z+x)
=y[(x+z)(x-z+y)]-zx(z+x)
=(x+z)[y(x-z+y-zx)]
=(x+z)(xy-yz+y2-xyz)
NN
0
a/ \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)
b/ \(\left(1-y\right)\left(y-x\right)\)
a. \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)
b. \(\left(1-y\right)\left(y-x\right)\)