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Ta có : \(A=\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\)
=> \(5A=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\)
Lấy 5A trừ A theo vế ta có :
5A - A = \(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{n}{5^n}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{n}{5^{n+1}}+...+\frac{11}{5^{12}}\right)\)
4A = \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)-\frac{11}{5^{12}}\)
Đặt B = \(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
=> 5B = \(1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
Lấy 5B trừ B ta có :
=> 5B - B = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\right)\)
=> 4B =\(1-\frac{1}{5^{11}}\)
=> B = \(\frac{1}{4}-\frac{1}{5^{11}.4}\)
Khi đó 4A = \(\frac{1}{4}-\frac{1}{5^{11}.4}-\frac{1}{5^{12}}\)
=> A = \(\frac{1}{16}-\left(\frac{1}{5^{11}.16}+\frac{1}{5^{12}.4}\right)< \frac{1}{16}\left(\text{ĐPCM}\right)\)
cậu ơi , mình quên không ghi 1 dữ liệu ạ
n thuộc N
V ậy có cần phải chỉnh sửa ở trong bài làm không ạ?????
a, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}.\frac{12}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=\frac{-5}{7}.1+\frac{12}{7}=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
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
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: \(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(=7+\frac{5}{11}-2-\frac{3}{7}-3-\frac{5}{11}\)
\(=2-\frac{3}{7}=\frac{11}{7}\)
\(\Rightarrow5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow5H-H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)\)
\(\Rightarrow4H=\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...+\frac{1}{5^{10}}\)
\(\Rightarrow5A-A=\left(1+..+\frac{1}{5^{10}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{11}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{11}}\)
\(\Rightarrow A=\frac{1}{4}-\frac{1}{4.5^{11}}\)
\(\Rightarrow4H=\frac{1}{4}-\frac{1}{4.5^{11}}-\frac{11}{5^{12}}\)
\(\Rightarrow H=\frac{1}{16}-\frac{1}{4^2.5^{11}}-\frac{11}{4.5^{12}}\)
Ta có : \(5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow4H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}+\frac{11}{5^{12}}\)
\(\Rightarrow20H=1+\frac{1}{5}+...+\frac{1}{5^{10}}+\frac{11}{5^{11}}\)
\(\Rightarrow16H=20H-4H=1+\frac{10}{5^{11}}-\frac{11}{5^{12}}\Leftrightarrow H=\frac{1+\frac{10}{5^{11}}-\frac{11}{5^{12}}}{16}.\)