Xét tam giác \(ABD\)và tam giác \(ACD\):

\(AB=AC\)(giả thiết) 

\(BD=CD\)(vì \(D\)là trung điểm của \(BC\))

\(AD\)cạnh chung

Suy ra \(\Delta ABD=\Delta ACD\left(c.c.c\right)\).

chia trước cộng sau thôi

chia trước cộng sau, theo quy tắc đó thì ta có:

   1+100000:12

= 1+ 25000/3

=  25003/3                             

 gạch "/ " là phần nha

llllllff,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,4485 1

/b;nlb, mkjv mmjg

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)

\(=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

\(\Rightarrow bz-cy=cx-az=ay-bx=0\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

\(B=2016:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}\right)\cdot\left(\frac{-1\frac{1}{6}+0,875-0,7}{\frac{1}{3}-0,25+\frac{1}{5}}\right)\)

\(B=2016:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\right)\cdot\left(\frac{-\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right)\)

\(B=2016:\frac{2}{7}\cdot\left(\frac{-7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}\right)\)

\(B=2016:\frac{2}{7}\cdot\left(\frac{-7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right)\)

\(B=2016:\frac{2}{7}\cdot\left(-\frac{7}{2}\right)\)

\(B=-24696\)