\(x^2-3x+xy-3y\\=(x^2-3x)+(xy-3y)\\=x(x-3)+y(x-3)\\=(x-3)(x+y)\)

sos

\(-x^2+6x+6y+y^2\\=-(x^2-6x+9)+(y^2+6y+9)\\=-(x-3)^2+(y+3)^2\\=(y+3)^2-(x-3)^2\\=[(y+3)-(x-3)][(y+3)+(x-3)]\\=(y+3-x+3)(y+3+x-3)\\=(y-x+6)(x+y)\)

\(225-4x^2-4xy-y^2\\=225-(4x^2+4xy+y^2)\\=225-[(2x)^2-2\cdot2x\cdot y+y^2]\\=15^2-(2x-y)^2\\=[15-(2x-y)][15+(2x-y)]\\=(15-2x+y)(15+2x-y)\)

Giúp vs m.n ơi mai mình kt òi

a) Với m=0

=> pt <=> \(x^2+5x=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b) \(x^2+5x+3m=0\)

\(\Delta=25-12m\)

Để phương trình có 2 nghiệm phân biệt 

\(\Leftrightarrow\Delta>0\)

\(\Leftrightarrow25-12m>0\)

\(\Leftrightarrow m< \dfrac{25}{12}\)

\(x\left(y^2+z^2\right)+y\left(z^2+x^2\right)+z\left(x^2+y^2\right)+2xyz=xy^2+xz^2+yz^2+x^2y++zx^2+zy^2+2xyz=xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)=\left(x+y+z\right)\left(xy+yz\right)+xz\left(x+z\right)=y\left(x+y+z\right)\left(z+x\right)+xz\left(x+z\right)=\left(x+z\right)\left(xy+y^2+yz+xz\right)=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)