a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=1-1+\dfrac{1}{72}\)

\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)

\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)

\(=-\left(-\dfrac{173}{1287}\right)\)

\(=\dfrac{173}{1287}\)

c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{-49}{50}\)

\(\dfrac{-7}{9}+\dfrac{-2}{9}=\dfrac{-9}{9}=-1\)

\(\dfrac{15}{4}-\dfrac{-3}{4}=\dfrac{18}{4}=\dfrac{9}{2}\)

\(\dfrac{1}{6}-\dfrac{1}{36}=\dfrac{6}{36}-\dfrac{1}{36}=\dfrac{5}{36}\)

\(1-\dfrac{3}{2}=\dfrac{2}{2}-\dfrac{3}{2}=\dfrac{-1}{2}\)

\(2-\dfrac{4}{5}=\dfrac{10}{5}-\dfrac{4}{5}=\dfrac{6}{5}\)

A = \(\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\)\(\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\)\(\dfrac{1}{72}\)

= \(1-1+\dfrac{7}{12}\)

=\(\dfrac{1}{72}\)

A = ( 1/3+3/5+1/15)-(3/4+2/9+1/36)+1/72

= 1-1+1/72

= 1/72

A=\(\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)

\(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{75}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{75}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{75}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{75}\)
\(=1-1+\dfrac{1}{75}\)
\(=0+\dfrac{1}{75}\)
\(=\dfrac{1}{75}\)
#AvoidMe

=1/3+3/5+1/15-3/4-2/9-1/36+1/75

=5/15+9/15+1/15-27/36-8/36-1/36+1/75

=1/75

\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)

=1/64

\(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{15}-\dfrac{2}{9}-\dfrac{1}{36}\)

\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)\)

=0

A \(=\)\(\dfrac{1}{3}\) \(-\) \(\dfrac{3}{4}\) \(+\) \(\dfrac{3}{5}\)\(+\) \(\dfrac{1}{15}\) \(-\) \(\dfrac{2}{9}\) \(-\) \(\dfrac{1}{36}\)                                                                       \(=\) \(\dfrac{1}{36}\) xong ùi nha

\(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{15}-\dfrac{2}{9}-\dfrac{1}{36}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{2}{9}-\dfrac{1}{36}\)

\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}\)

=0