bài 1
a)(x-1)(x+4)
= $x^{2}$+4x-x-4
= $x^{2}$+3x-4
b)$\frac{3}{2x-6}$-$\frac{x+6}{2x^2-6x}$
=$\frac{3}{2(x-3)}$- $\frac{x+6}{2x(x-3)}$
=$\frac{3x}{2x(x-3)}$- $\frac{x+6}{2x(x-3)}$
=$\frac{3x-x-6}{2x(x-3)}$=$\frac{2x-6}{2x(x-3)}$
=$\frac{2(x-3)}{2x(x-3)}$
= $\frac{1}{x}$
bài 2:
a)$x^{2}$-$y^{2}$+5x+5y
=(x-y)(x+y)+5(x+y)
=(x+y)(x-y+5)
b)xy-$y^{2}$+5y-5x
=y(x-y)+5(y-x)
=y(x-y)-5(x-y)
=(x-y)(y-5)
bài 3:
a)$x^{2}$+2x=0
⇔x(x+2)=0
⇔\(\left[ \begin{array}{l}x=0\\x+2=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
Vậy x=0 hoặc x=-2
b)$x^{3}$-4x=0
⇔x($x^{2}$-4)=0
⇔x(x-2)(x+2)=0
TH1:x=0
TH2:x-2=0
⇒x=2
TH3:x+2=0⇒x=-2
Vậy=0 hoặc x=-2 hoặc x=2
chúc bạn học tốt ^-^