a: \(\dfrac{-2}{x^2+6x}=\dfrac{-2}{x\left(x+6\right)}\)

\(\dfrac{x}{x+6}=\dfrac{x\cdot x}{x\left(x+6\right)}=\dfrac{x^2}{x\left(x+6\right)}\)

b: \(\dfrac{11}{25-x^2}=\dfrac{-11}{x^2-25}=\dfrac{-11}{\left(x-5\right)\left(x+5\right)}=\dfrac{-11\left(x-5\right)}{\left(x-5\right)^2\cdot\left(x+5\right)}\)

\(\dfrac{2}{x^2-10x+25}=\dfrac{2}{\left(x-5\right)^2}=\dfrac{2\left(x+5\right)}{\left(x+5\right)\cdot\left(x-5\right)^2}\)

Bài 2: 

Xét ΔBDC có BE/ED=BF/FC

nên EF//DC

mà DC//AB

nên EF//AB

b3:

Bảo toàn cơ năng:

m.g.h=\(\dfrac{1}{2}.m.\left(vđất\right)^2\)

\(=>v_{đất}=8\sqrt{2}\)

Hình như hình kẻ bị sai

a) Ta có: a//b,a⊥c

=> b⊥c

b) Kẻ Oz//a//b(Oz thuộc một nửa mặt phẳng bờ OB không chứa điểm D

\(\Rightarrow\left\{{}\begin{matrix}\widehat{AOz}=\widehat{OAa}=40^0\\\widehat{BOz}=\widehat{OBb}=50^0\end{matrix}\right.\)(so le trong)

\(\Rightarrow\widehat{AOB}=\widehat{AOz}+\widehat{BOz}=40^0+50^0=90^0\)

\(\frac{2}{5}x+\frac{1}{3}x=\frac{2}{3}\Leftrightarrow\frac{11}{15}x=\frac{2}{3}\Leftrightarrow x=\frac{2}{3}:\frac{11}{15}\Leftrightarrow x=\frac{10}{11}\)

C1 : 2/5 . x + 1/3 . x = 2/3

       x . ( 2/5 + 1/3 )  = 2/3

       x . 11/15           = 2/3

           x                  = ( 2/3 ) / ( 11/15 )

           x                  = 10/11

Vậy x = 10/11

c2: mk có chơi nè 

16384/25 =655.36

\(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{7}{20}=\dfrac{12}{20}-\dfrac{5}{20}+\dfrac{7}{20}=\dfrac{14}{20}=\dfrac{7}{10}\)

\(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{7}{20}=\dfrac{12}{20}-\dfrac{5}{20}+\dfrac{7}{20}=\dfrac{7}{20}+\dfrac{7}{20}=\dfrac{14}{20}\)

`3/5-1/4+7/20=12/20-5/20+7/20=14/20=7/10`