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Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)
\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)
\(\Rightarrow6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7A=(1+\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}})-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6A=\left(1-\frac{1}{7^{99}}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{7^{99}}\right):6\)
Câu b tương tự nha
a) \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...........+\frac{1}{7^{100}}\)
\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.........+\frac{1}{7^{99}}\)
\(\Rightarrow7A-A=6A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)
Ta có: \(\frac{3}{5}+\frac{-1}{3}=\frac{4}{15}\)
Số đối của \(\frac{4}{15}\) là \(\frac{-4}{15}\)
\(\frac{-2}{13}+\frac{-11}{26}=\frac{-15}{26}\)
Số đối của \(\frac{-15}{26}\) là \(\frac{15}{26}\)
\(-2+\frac{-5}{8}=\frac{-21}{8}\)
Số đối của \(-\frac{21}{8}\) là \(\frac{21}{8}\)
\(2+\frac{-3}{4}=\frac{5}{4}\)
Số đối của \(\frac{5}{4}\) là \(\frac{-5}{4}\)
\(\frac{13}{3}+\frac{5}{3}=6\)
Số đối của 6 là -6
\(\frac{-1}{3}+\frac{7}{3}=2\)
Số đối của 2 là -2
\(\frac{-7}{2}+\frac{-3}{4}=\frac{-17}{4}\)
Số đối của \(\frac{-17}{4}\) là \(\frac{17}{4}\)
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
Bài 1:
a) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=4\frac{5}{7}-1\frac{3}{4}\)
\(=\frac{33}{7}-\frac{7}{4}\)
\(=\frac{132}{28}-\frac{49}{28}=\frac{83}{28}\)
b) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=4\frac{5}{9}-2\frac{3}{4}\)
\(=\frac{41}{9}-\frac{11}{4}\)
\(=\frac{164}{36}-\frac{99}{36}=\frac{65}{36}\)
c) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\cdot\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=\frac{-3}{5}\cdot2=-\frac{6}{5}\)
d) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) \(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}\)
\(=\left(6\frac{4}{5}-3\frac{4}{5}\right)-1\frac{2}{3}\)
\(=\left(6+\frac{4}{5}-3-\frac{4}{5}\right)-1\frac{2}{3}\)
\(=3-\frac{5}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)
h) \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=\left(7\frac{5}{9}-3\frac{5}{9}\right)-2\frac{3}{4}\)
\(=\left(7+\frac{5}{9}-3-\frac{5}{9}\right)-2\frac{3}{4}\)
\(=4-\frac{11}{4}=\frac{16}{4}-\frac{11}{4}=\frac{5}{4}\)
i) \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=\left(6+\frac{5}{7}-2-\frac{5}{7}\right)-\frac{7}{4}\)
\(=4-\frac{7}{4}=\frac{16}{4}-\frac{7}{4}=\frac{9}{4}\)
k) \(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(=7\frac{5}{11}-2\frac{3}{7}-3\frac{5}{11}\)
\(=\left(7\frac{5}{11}-3\frac{5}{11}\right)-2\frac{3}{7}\)
\(=4-\frac{17}{7}=\frac{28}{7}-\frac{17}{7}=\frac{11}{7}\)
g) Ta có: \(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=\frac{34}{5}-\left(\frac{5}{3}+\frac{19}{5}\right)\)
\(=\frac{34}{5}-\frac{5}{3}-\frac{19}{5}\)
\(=3-\frac{5}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)
h) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=\frac{68}{9}-\frac{11}{4}-\frac{32}{9}\)
\(=4-\frac{11}{4}=\frac{16}{4}-\frac{11}{4}=\frac{5}{4}\)
i) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6+\frac{5}{7}-1-\frac{3}{4}-2-\frac{5}{7}\)
\(=3-\frac{3}{4}=\frac{9}{4}\)
k) Ta có: \(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(=7+\frac{5}{11}-2-\frac{3}{7}-3-\frac{5}{11}\)
\(=2-\frac{3}{7}=\frac{11}{7}\)
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)