bài 1
a,2x(x-3)-x(x+5)
=x(2x-6)-x(x+5)
=x(x-11)
b,$\frac{10x^2z^3y^4}{15x^3y^2z^2}$ =$\frac{2zy^2}{3x}$
d,$\frac{x^2+5x+6}{x^2+7x+10}$ =$\frac{(x+2)(x+3)}{(x+2)(x+5)}$ =$\frac{x+3}{x+5}$
c;$\frac{16-x^2}{x^2+8x+16}$ =$\frac{(4+x)(4-x)}{(x+4)^2}$ =$\frac{4-x}{4+x}$
bài 2
a,2$x^{2}$-6x=0
⇔2x(x-3)=0
⇔\(\left[ \begin{array}{l}2x=0\\x-3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\)
b,$x^{2}$ +7x+12=0
⇔$x^{2}$ +3x+4x+12=0
⇔x(x+3)+4(x+3)=0
⇔(x+3)(x+4)=0
⇔\(\left[ \begin{array}{l}x+3=0\\x+4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=-4\end{array} \right.\)
c,$x^{2}$ +10x+23=0
⇔(x-$\sqrt{2}$ +5)(x+$\sqrt{2}$ +5)=0⇔ \(\left[ \begin{array}{l}x=-5+căn2\\x=-5-căn2\end{array} \right.\)
