Bài 1 1) Phân tích đa thức thành nhân tử: a)x^2-5xy b)x^2-y^2+5x-5y c)x^2+2xy+y^2-16 2)Tìm x biết:2x(x-5)=4(x-5)

$a, x^2-5xy=x(x-5y)$

$b, x^2-y^2+5x-5y$

$=(x-y)(x+y)+5(x-y)$

$=(x-y)(x+y+5)$

$c, x^2+2xy+y^2-16$

$=(x+y)^2-4^2$

$=(x+y-4)(x+y+4)$

$2) 2x(x-5)=4(x-5)$

$=>2x(x-5)-4(x-5)=0$

$=>(x-5)(2x-4)=0$

$=>\(\left[ \begin{array}{l}x-5=0\\2x-4=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=5\\2x=4\end{array} \right.\) =>\(\left[ \begin{array}{l}x=5\\x=2\end{array} \right.\) $

$Vậy x=5, x=2$

@(Học tốt nhé!)@