Phân tích thành nhân tử: A = (xy + yz + zx) (x + y+ z) – xyz
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YY
1
10 tháng 10 2017
Ta có :
A=xy(x+y+z) -xyz +(yz+zx)(x+y+z)
A=xy(x+y+z -z) + z(x+y)(x+y+z)
A=xy(x+y) +z(x+y)(x+y+z)
A=(x+y)(xy+z(x+y+z))
A=(x+y)(xy+zx+z(y+z))
A=(x+y)(x(y+z)+z(y+z))
A=(x+y)(y+z)(x+z)
P/ s : ko biết có đúng ko
NM
1
30 tháng 7 2015
xy( x+ y) + yz(y+z) + xz(x+z) + 3xyz
= xy(x+y) + xyz + yz(y+z) + xyz + xz(x+z) + xyz
= zy(x+y+z) + yz(x + y + z) + xz ( x+y+z)
= ( x+ y +z )( xy + yz + zx)
BC
1
30 tháng 9 2015
A=x2y+xy2+xyz+xyz+y2z+yz2+x2z+xyz+xz2-xyz
A=(x2y+xy2+xyz+y2z)+(yz2+x2z+xyz+xz2)
A=y(x2+xy+xz+yz)+z(yz+x2+xy+xz)
A=(y+z)(x2+xy+xz+yz)
A=(y+z)[x(x+y)+z(x+y)]
A=(y+z)(x+y)(x+z)
6 tháng 1 2018
(xy+yz+zx)(x+y+z)-xyz
= (xy+yz)(x+y+z)+ x2z+xyz+ xz2-xyz
= (xy+yz)(x+y+z)+ x2z + xz2
= (x+z)(xy+y2+zy)+ xz(x+z) = (x+z)(xy+y2+zy+xz) = (x+z)(x+y)(x+z)
TH
1
CM
17 tháng 4 2019
Ta có
C = xyz – (xy + yz + zx) + x + y + z – 1
= (xyz – xy) – (yz – y) – (zx – x) + (z – 1)
= xy(z – 1) – y(z – 1) – x(z – 1) + (z – 1)
= (z – 1)(xy – y – x + 1)
= (z – 1).[y(x – 1) – (x – 1)]
= (z – 1)(y – 1)(x – 1)
Với x = 9; y = 10; z = 101 ta có
C = (101 – 1)(10 – 1)(9 – 1) = 100.9.8 = 7200
Đáp án cần chọn là: C
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. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)
22 tháng 10 2021
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
Ta có
A=(x+y)(xy+yz+xz)+z(xy+yz+xz)-xyz
=(x+y)(xy+yz+xz)+xyz+z^2(y+x)-xyz
= (x+y)(xy+yz+xz + z^2)
= (x+y)(y+z)(x+z)
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