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\(\frac{3}{5}+\frac{3}{5^3}+\frac{3}{5^5}+...+\frac{3}{5^{199}}\)
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\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
a, Dat A =\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\)
\(\Rightarrow\frac{1}{3}A+A=\left(\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\right)+\left(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\right)\)
\(\Rightarrow\frac{4}{3}A=\frac{1}{3}+\frac{1}{3^{200}}\)
\(\Rightarrow A=\frac{\frac{1}{3}+\frac{1}{3^{200}}}{\frac{4}{3}}\)
chung minh tuong tu cau b va c
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)
Bài làm ai trên 11 điểm tích mình thì mình tích lại
Ông tùng hơn tùng số tuổi là :
29 + 32 = 61 (tuổi )
Vậy ông của tùng hơn tùng 61 tuổi
(A) \(\frac{2}{3} + \frac{{ - 4}}{6} = \frac{4}{6} + \frac{{ - 4}}{6} = 0\) => A sai
(B) \(\frac{2}{3}.\frac{{ - 1}}{5} = \frac{{ - 2}}{{15}}\) mà \(\frac{{3 - 2}}{5} = \frac{1}{5}\) => B sai
(C) \(\frac{2}{3} - \frac{3}{5} = \frac{{10}}{{15}} - \frac{9}{{15}} = \frac{1}{{15}}\) => C đúng
(D) \(\frac{3}{5}:\frac{3}{{ - 5}} = \frac{3}{5}.\frac{{ - 5}}{3} = \frac{{ - 15}}{{15}} = - 1\) => D sai
=> Chọn C.
a)
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}}:\frac{3}{{ - 4}}} \right).\frac{4}{5} = \left( {\frac{2}{5}.\frac{{ - 4}}{3}} \right).\frac{4}{5}\\ = \frac{{ - 8}}{{15}}.\frac{4}{5} = \frac{{ - 32}}{{75}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}}:\left( {\frac{7}{{ - 5}}.\frac{{ - 3}}{2}} \right) = \frac{3}{4}:\frac{{ - 21}}{{ - 10}}\\ = \frac{3}{4}.\frac{{10}}{{21}} = \frac{{30}}{{84}} = \frac{5}{14}\end{array}\)
c)
\(\begin{array}{l}\frac{{ - 1}}{9}.\frac{{ - 3}}{5} + \frac{5}{{ - 6}}.\frac{{ - 3}}{5} + \frac{5}{2}.\frac{{ - 3}}{5}.\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 1}}{9} + \frac{5}{{ - 6}} + \frac{5}{2}} \right)\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 15}}{{18}} + \frac{{45}}{{18}}} \right)\\ = \frac{{ - 3}}{5}.\frac{{28}}{{18}}\\ = \frac{{ - 3}}{5}.\frac{{14}}{9}\\ = \frac{{ - 14}}{{15}}\end{array}\)
\(\left(\frac{3}{5}-\frac{1}{2}+\frac{7}{3}\right)-\left(\frac{1}{3}-\frac{5}{2}+\frac{1}{5}\right)-\left(-\frac{3}{5}+3-\frac{5}{3}\right)\)
\(=\frac{3}{5}-\frac{1}{2}+\frac{7}{3}-\frac{1}{3}+\frac{5}{2}-\frac{1}{5}+\frac{3}{5}-3+\frac{5}{3}\)
\(=\frac{3}{5}+\frac{-1}{2}+\frac{7}{3}+\frac{-1}{3}+\frac{5}{2}+\frac{-1}{5}+\frac{3}{5}+\left(-3\right)+\frac{5}{3}\)
\(=\left(\frac{3}{5}+\frac{-1}{5}+\frac{3}{5}\right)+\left(\frac{-1}{2}+\frac{5}{2}\right)+\left(\frac{7}{3}+\frac{-1}{3}+\frac{5}{3}\right)+\left(-3\right)\)
\(=\frac{5}{2}+\frac{4}{2}+\frac{11}{3}+\left(-3\right)\)
\(=1+2+\frac{11}{3}+\left(-3\right)\)
\(=0+\frac{11}{3}\)
\(=\frac{11}{3}\)
\(\left(\frac{3}{5}-\frac{1}{2}+\frac{7}{3}\right)-\left(\frac{1}{3}-\frac{5}{2}+\frac{1}{5}\right)-\left(-\frac{3}{5}+3-\frac{5}{3}\right)\)
\(=\frac{3}{5}-\frac{1}{2}+\frac{7}{3}-\frac{1}{3}+\frac{5}{2}-\frac{1}{5}+\frac{3}{5}-3+\frac{5}{3}\)
\(=\left(\frac{3}{5}-\frac{1}{5}+\frac{3}{5}\right)+\left(-\frac{1}{2}+\frac{5}{2}\right)+\left(\frac{7}{3}-\frac{1}{3}+\frac{5}{3}\right)-3\)
\(=1+2+\frac{11}{3}-3\)
\(=\left(1+2-3\right)+\frac{11}{3}\)
\(=0+\frac{11}{3}\)
=\(\frac{11}{3}\)
(Bài làm có gì ko hiểu bn cứ hỏi mk nhé ^...^ ^_^)
\(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{4\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{8}\right)}{5\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{8}\right)}\)
\(=\frac{3}{5}+\frac{4}{5}=\frac{7}{5}\)
Ta có: \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{4\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{8}\right)}{5\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{8}\right)}\)
\(=\frac{3}{5}+\frac{4}{5}\)
\(=\frac{7}{5}\)
\(A=\frac{3}{5}+\frac{3}{5^3}+\frac{3}{5^5}+...+\frac{3}{5^{199}}\)
\(25\text{A}=15+\frac{3}{5}+\frac{3}{5^3}+...+\frac{3}{5^{197}}\)
\(25\text{A}-A=\left(15+\frac{3}{5}+\frac{3}{5^3}+...+\frac{3}{5^{197}}\right)-\left(\frac{3}{5}+\frac{3}{5^3}+\frac{3}{5^5}+...+\frac{3}{5^{199}}\right)\)
\(24A=15-\frac{3}{5^{199}}\)
\(A=\frac{\left(15-\frac{3}{5^{199}}\right)}{24}\)