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

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

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

\(\Rightarrow A=1+\left(-1\right)+\frac{1}{72}\)

\(\Rightarrow A=0+\frac{1}{72}\)

\(\Rightarrow A=\frac{1}{72}\)

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

đúng rồi ! OK!

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

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

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

Rất vui vì giúp đc ban <3

\(A=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{72}\\ =\frac{1}{9}-\frac{7}{9}+\frac{2}{3}+\frac{1}{72}=-\frac{2}{3}+\frac{2}{3}+\frac{1}{72}=\frac{1}{72}\)

\(\frac{1}{72}\)

nha bn 

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

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

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

= 1/72

Đảm bảo 100 % là đúng

C = 1/100 - ( 1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99

C = 1/100 - (  1- 1/2+ 1/2 - 1/3 + ... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100 )

C = 1/100 - ( 1 - 1/100 )

C = 1/100 - 99/100

C = \(\frac{-49}{50}\)