`a)x^2+4x`
`=x(x+4)`
`b)5(x-y)-y(x-y)`
`=(5-y)(x-y)`
`c)x^2+2x+1-y^2`
`=(x^2+2x+1)-y^2`
`=(x^2+2.x.1+1^2)-y^2`
`=(x+1)^2-y^2`
`=(x+1+y)(x+1-y)`
***
`a)5(x-1)+10=0`
`⇒5x-5+10=0`
`⇒5x+5=0`
`⇒5x=-5`
`⇒x=-1`
Vậy `x=-1`
`b)2x(x+5)-x-5=0`
`⇒2x(x+5)-(x+5)=0`
`⇒(2x-1)(x+5)=0`
`⇒` \(\left[ \begin{array}{l}2x-1=0\\x+5=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=-5\end{array} \right.\)
Vậy `x∈{1/2;-5}`