Hình tự vẽ, Giải : 

Kẻ tia phân giác góc A => góc A1 = góc A2. 

Tia phân giác góc A cắt BC tại M

Tự các dữ kiện suy ra tam giác ABM = tam giác ACM ( c.c.c ) 

Suy ra góc B = góc C ( tương ứng ) 

A I B N M C

TA CÓ:

IM là cạnh chung

BI=MN(gt)

góc MIB=góc IMN  (AB//MN)

TAM giác IBM=Tam giác INM(c-g-c)

góc BMI=góc MIN

suy ra IM//AC

a)\(\frac{7}{12}.\frac{6}{11}+\frac{7}{12}.\frac{5}{11}-2\frac{7}{12}\)

\(=\frac{7}{12}.\left(\frac{6}{11}+\frac{5}{11}\right)-\frac{31}{12}\)

\(=\frac{7}{12}-\frac{31}{12}\)

\(=-2\)

b)\(\frac{-5}{9}.\frac{-6}{13}+\frac{5}{-9}.\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{5}{9}.\left(\frac{6}{13}+\frac{5}{13}-1\right)\)

\(=\frac{5}{9}.\left(\frac{11}{13}-\frac{13}{13}\right)\)

\(=\frac{5}{9}.\frac{-2}{13}\)

\(=-\frac{10}{117}\)

c)\(0,8.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}.\left(-\frac{15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}.\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

d)\(-75\%.\frac{6}{7}+5\%.\frac{6}{7}+\frac{7}{10}.1\frac{1}{7}\)

\(=\frac{-15}{20}.\frac{6}{7}+\frac{1}{20}.\frac{6}{7}+\frac{7}{10}.\frac{8}{7}\)

\(=\frac{6}{7}.\left(\frac{-15}{20}+\frac{1}{20}\right)+\frac{4}{5}\)

\(=\frac{6}{7}.\frac{-7}{10}+\frac{4}{5}\)

\(=-\frac{3}{5}+\frac{4}{5}\)

\(=\frac{1}{5}\)

Linz

\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+3^{10}.2^{20}}\)

\(=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)

\(M=\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\) 

\(=\frac{2^{19}\cdot\left(3^3\right)^3+3\cdot5\cdot\left(2^2\right)^9\cdot\left(3^2\right)^4}{6^9\cdot2^{10}+6^{10}\cdot2^{10}}\) 

\(=\frac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3\cdot3^8}{6^9\cdot2^{10}\left(6+1\right)}\) 

\(=\frac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3^9}{6^9\cdot2^{10}\cdot7}\) 

\(=\frac{2^{18}\cdot3^9\left(2+5\right)}{2^{10}\cdot2^9\cdot3^9\cdot7}\) 

\(=\frac{2^{18}\cdot3^9\cdot7}{2^{19}\cdot3^9\cdot7}\) 

\(=\frac{1}{2}\)

lếu lều ko nếu có thì ib mik nhá

\(\frac{9}{8}-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{72}=\frac{9}{8}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)

\(=\frac{9}{8}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)

\(=\frac{9}{8}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{9}{8}-\left(1-\frac{1}{9}\right)=\frac{9}{8}-\frac{8}{9}=\frac{17}{72}\)