Toán `xy(x+y)-yz(y+z)+xz(x-z)`Phân tích thành nhân tử 26/07/2021 By Delilah `xy(x+y)-yz(y+z)+xz(x-z)`Phân tích thành nhân tử
Đáp án: `xy(x+y)-yz(y+z)+xz(x-z)` `=xy(x-z+z+y)-yz(z+y)+xz(x-z)` `=xy(x-z)+xy(z+y)-yz(x+y)+xz(x-z)` `=[xy(x-z)+xz(x-z)]+[xy(z+y)-yz(x+y)]` `=(x-z)(xy+xz)+(z+y)(xy-yz)` `=x(x-z)(y+z)+y(y+z)(x-z)` `=(x-z)(y+z)(x+y)` Trả lời
xy(x+y)−yz(y+z)+xz(x−z) =y.[x(z+y)+z(y+z)]+zx(x-z) =y.(xz+xy+zy+z2)+zx(x-z) =y.(xz+z2+xy+zy)+zx(x-z) =y.[z.(z+x)+y.(z+x)]+zx(x-z) =y.(z+x)(z+y)-zx(x+z) =(z+x)[y(z+y)-zx] =(z+x)(yz+y^2-zx) Trả lời
Đáp án:
`xy(x+y)-yz(y+z)+xz(x-z)`
`=xy(x-z+z+y)-yz(z+y)+xz(x-z)`
`=xy(x-z)+xy(z+y)-yz(x+y)+xz(x-z)`
`=[xy(x-z)+xz(x-z)]+[xy(z+y)-yz(x+y)]`
`=(x-z)(xy+xz)+(z+y)(xy-yz)`
`=x(x-z)(y+z)+y(y+z)(x-z)`
`=(x-z)(y+z)(x+y)`
xy(x+y)−yz(y+z)+xz(x−z)
=y.[x(z+y)+z(y+z)]+zx(x-z)
=y.(xz+xy+zy+z2)+zx(x-z)
=y.(xz+z2+xy+zy)+zx(x-z)
=y.[z.(z+x)+y.(z+x)]+zx(x-z)
=y.(z+x)(z+y)-zx(x+z)
=(z+x)[y(z+y)-zx]
=(z+x)(yz+y^2-zx)