Toán Phần tích đã thức thành nhân tử xy(x+y) +yz(y+z) +xz(x+z) +2xyz 04/07/2021 By Brielle Phần tích đã thức thành nhân tử xy(x+y) +yz(y+z) +xz(x+z) +2xyz
`xy(x+y) +yz(y+z) +xz(x+z) +2xyz` `=x^2y+xy^2+y^2z+yz^2+xz(x+z)+2xyz` `=(x^2y+xyz)+(xy^2+y^2z)+(yz^2+xyz)+xz(x+z)` `=xy(x+z)+y^2(x+z)+yz(x+z)+xz(x+z)` `=(x+z)(xy+y^2+yz+xz)` `=(x+z)[y(x+y)+z(x+y)]` `=(x+z)(x+y)(y+z)` Trả lời
Đáp án: $xy(x + y) + yz(y + z) + xz(x + z) + 2xyz$ $= xy(x + y) + yz(y + z) + xzy + xz(x + z) + xzy$ $= xy(x + y) + yz(y + z + x) + xz(x + z + y)$ $= xy(x + y) + z(x + y + z)(x + y)$ $= xy(x + y) + (xz + zy + z²)(x + y)$ $= (x + y)(xy + xz + zy + z²)$ $= (x + y)[x(y + z) + z(y + z)]$ $= (x + y)(y + z)(z + x) $ Trả lời
`xy(x+y) +yz(y+z) +xz(x+z) +2xyz`
`=x^2y+xy^2+y^2z+yz^2+xz(x+z)+2xyz`
`=(x^2y+xyz)+(xy^2+y^2z)+(yz^2+xyz)+xz(x+z)`
`=xy(x+z)+y^2(x+z)+yz(x+z)+xz(x+z)`
`=(x+z)(xy+y^2+yz+xz)`
`=(x+z)[y(x+y)+z(x+y)]`
`=(x+z)(x+y)(y+z)`
Đáp án:
$xy(x + y) + yz(y + z) + xz(x + z) + 2xyz$
$= xy(x + y) + yz(y + z) + xzy + xz(x + z) + xzy$
$= xy(x + y) + yz(y + z + x) + xz(x + z + y)$
$= xy(x + y) + z(x + y + z)(x + y)$
$= xy(x + y) + (xz + zy + z²)(x + y)$
$= (x + y)(xy + xz + zy + z²)$
$= (x + y)[x(y + z) + z(y + z)]$
$= (x + y)(y + z)(z + x) $