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\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{2004.2005}\)
\(\Leftrightarrow2M=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{2004.2005}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{2004.2005}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2004}-\frac{1}{2005}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2005}\right)\)
\(=2.\left(\frac{2005}{4010}-\frac{2}{4010}\right)\)
\(=2.\frac{2003}{4010}\)
\(=\frac{2003}{2005}\)
\(M=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{2004\cdot2005}\)
\(M=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2004\cdot2005}\)
\(M=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2004\cdot2005}\right)\)
\(M=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2004\cdot2005}\right)\)
\(M=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2004}-\frac{1}{2005}\right)\)
\(M=2\left(\frac{1}{2}-\frac{1}{2005}\right)\)
\(M=2\cdot\frac{2003}{4010}\)
\(M=\frac{2003}{2005}\)
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}\right)=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{9.10}\right)\)
\(=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...-\frac{1}{10}\right)=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{1}{5}-\frac{1}{4}=\frac{-1}{20}\)
\(A=\frac{1}{10}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\)
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{1}{10}-\left[\left(\frac{1}{4}-\frac{1}{10}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-...-\left(\frac{1}{9}-\frac{1}{9}\right)\right]\)
\(A=\frac{1}{10}-\frac{1}{4}+\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{4}\)
\(A=-\frac{1}{20}\)
1) \(\frac{3^{10}+6^2}{5\cdot3^8+20}=\frac{3^{10}+3^2\cdot2^2}{5\cdot3^8+5\cdot2^2}=\frac{3^2\left(3^8+2^2\right)}{5\left(3^8+2^2\right)}=\frac{9}{5}\)
2) \(\frac{28^{15}\cdot3^{17}}{84^{16}}=\frac{28^{15}\cdot3^{17}}{28^{16}\cdot3^{16}}=\frac{3}{28}\)
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).......\left(\frac{1}{98}+1\right).\left(\frac{1}{99}+1\right)\)
\(T=\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right).....\left(\frac{1}{98}+\frac{98}{98}\right).\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{99}{98}.\frac{100}{99}\)
\(T=\frac{3.4.5....99.100}{2.3.4.....98.99}\)
\(T=\frac{100}{2}\)
\(T=50\)
Vậy T = 50
Chúc bạn học tốt!
a. \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\)
\(=\frac{1}{3}(\frac{4}{5}+\frac{6}{5})-\frac{5}{3}\)
\(=\frac{1}{3}.2-\frac{5}{3}\)
\(=\frac{2}{3}-\frac{5}{3}\)
\(=-\frac{1}{1}\)
c. \(\frac{6}{7}.\frac{10}{9}+\frac{1}{7}.\frac{10}{9}-\frac{8}{9}\)
\(=\frac{10}{9}\left(\frac{6}{7}+\frac{1}{7}\right)-\frac{8}{9}\)
\(=\frac{10}{9}.1-\frac{9}{8}\)
\(=\frac{10}{9}-\frac{9}{8}\)
\(=-\frac{1}{72}\)
a) Ta có: \(A=\frac{2^{2017}}{2^{2017}}+\frac{2^{2016}}{2^{2017}}+\frac{2^{2015}}{2^{2017}}+...+\frac{2^1}{2^{2017}}+\frac{1}{2^{2017}}\)
\(=\frac{1+2^1+2^2+...+2^{2016}+2^{2017}}{2^{2017}}\)
Đặt: B=\(1+2^1+2^2+...+2^{2017}\)
\(\Leftrightarrow2B=2^1+2^2+2^3+....+2^{2017}+2^{2018}\)
\(\Leftrightarrow2B-B=2^{2018}-1\)
\(\Leftrightarrow B=2^{2018}-1\)
\(\Rightarrow A=\frac{B}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
Mik chỉ biết làm phần a thôi
b/ Sử dụng quy tắc: \(\frac{a+c}{b+c}< \frac{a}{b}\) với \(\left\{{}\begin{matrix}a;b;c>0\\a>b\end{matrix}\right.\)
\(B=\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}-1+2}{2^{10}-3+2}=\frac{2^{10}+1}{2^{10}-1}\)
\(\Rightarrow B>A\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{2004.2005}\)
Ta có: \(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2004.2005}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{2004.2005}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2004.2005}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2005}\right)\)
\(A=\frac{2003}{2005}\)
bn ơi bn chưa nhân với 2