Question

Phân tích đa thức thành nhân tử

xy(x+y)+yz(y+z)+xz(x+z)+3xyz

Answer 1

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+3xyz\\ =xy\left(x+y\right)+xyz+yz\left(y+z\right)+xyz+xz\left(x+z\right)+xyz\\ =xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+y+z\right)\\ =\left(x+y+z\right)\left(xy+yz+xz\right)\)

Answer 2

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+3xyz\)

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

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

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

Answer 3

\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+3xyz\\ =\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+\left[xz\left(x+z\right)+xyz\right]\\ =xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z+y\right)\\ =\left(x+y+z\right)\left(xy+yz+xz\right)\)

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