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\(4A=\frac{4}{3.7}+...+\frac{4}{95.99}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)
\(4A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{8}{99}\)
\(A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}\)
\(A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\frac{32}{99}\)
\(A=\frac{8}{99}\)
A = 4/3 x 7 + 4/7 x 11 + 4/11 x 15 + .... + 4/95 x 99
A = 4/3 - 4/7 + 4/7 - 4/11 + 4/11 - 4/15 + ..... + 4/95 - 4/99
A = 4/3 - 4/99
A = 128/99
\(A=4\left(\frac{4}{3.7}+\frac{4}{7.11}+.......+\frac{4}{95.99}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.......+\frac{1}{95}-\frac{1}{99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4.\left(\frac{33}{99}-\frac{1}{99}\right)\)
\(=4.\frac{32}{99}=\frac{128}{99}\)
a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
a) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)
\(=\frac{15-4}{26}=\frac{11}{26}\)
c) \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}=\frac{-22-1}{23}=\frac{-23}{23}=-1\)
a. \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\frac{15}{13.2}-\frac{2}{13}=\frac{15}{13.2}-\frac{2.2}{13.2}=\frac{15-4}{26}=\frac{11}{26}\)
C. \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{1}{23}.\left(-11.\frac{6}{7}-11.\frac{8}{7}-1\right)=\frac{1}{23}.\left(-22-1\right)=\frac{1}{23}.\left(-23\right)=-1\)
Cây a, bạn nhân cả 2 vế với 3
Lấy vế nhân với 3 trừ đi ban đầu tất cả chia 2
b) Tính như bình thường
Câu c hình như sai đề
a=1/3+1/99=34/99
Ta có: \(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)