`1a)3x^3-6x^2`
`=3x^2(x-3)`
`b)x^2-2xy-25+y^2`
`=(x^2-2xy+y^2)-25`
`=(x-y)^2-5^2`
`=(x-y-5)(x-y+5)`
`c)3x^2-7x-10`
`=3x^2+3x-10x-10`
`=(3x^2+3x)-(10x+10)`
`=3x(x+1)-10(x+1)`
`=(3x-10)(x+1)`
`2)x^2-2xy+y^2+6y+10=0`
`→x^2-2xy+y^2+6y+1+9=0`
`→(x^2-2xy+1)+(y^2+6y+9)=0`
`→(x-1)^2+(y+3)^2=0`
`→` $\begin{cases}(x-1)^2=0\\(y+3)^2=0\\\end{cases}$
`→` $\begin{cases}x-1=0\\y+3=0\\\end{cases}$
`→` $\begin{cases}x=1\\y=-3\\\end{cases}$
Vậy `(x;y)=(1;-3)`