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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, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}.\frac{12}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=\frac{-5}{7}.1+\frac{12}{7}=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)
\(=-\frac{19}{8}\)
b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{5}{12}\)
c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)
\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)
\(=-\frac{43}{30}\)
d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)
\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)
\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)
\(=\frac{3}{20}\)
e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)
\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)
f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)
\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)
g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)
\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)
\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)
h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)
\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)
\(=\frac{78149}{6210}\)
i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)
\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)
\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)
\(=\frac{55}{4}\)
1: Ta có: \(\frac{12}{7}:\frac{16}{5}\)
\(=\frac{12}{7}\cdot\frac{5}{16}=\frac{60}{112}=\frac{15}{28}\)
2: Ta có: \(\frac{-2}{7}:\frac{6}{11}\)
\(=\frac{-2}{7}\cdot\frac{11}{6}=\frac{-22}{42}=\frac{-11}{21}\)
3: Ta có: \(\frac{-1}{7}:\frac{-1}{5}\)
\(=\frac{-1}{7}\cdot\frac{5}{-1}=\frac{5}{7}\)
4: Ta có: \(\left(-2\right):\frac{6}{11}\)
\(=\frac{-2\cdot11}{6}=\frac{-22}{6}=\frac{-11}{3}\)
5: Ta có: \(\frac{-6}{11}:\left(-3\right)\)
\(=\frac{-6}{11}\cdot\frac{1}{-3}=\frac{-6}{-33}=\frac{2}{11}\)
\(\Rightarrow5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow5H-H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)\)
\(\Rightarrow4H=\frac{1}{5}+\frac{1}{5^2}+..+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
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 : \(5H=\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\)
\(\Rightarrow4H=\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{11}{5^{12}}\right)=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{11}}+\frac{11}{5^{12}}\)
\(\Rightarrow20H=1+\frac{1}{5}+...+\frac{1}{5^{10}}+\frac{11}{5^{11}}\)
\(\Rightarrow16H=20H-4H=1+\frac{10}{5^{11}}-\frac{11}{5^{12}}\Leftrightarrow H=\frac{1+\frac{10}{5^{11}}-\frac{11}{5^{12}}}{16}.\)