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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 \(\left\{{}\begin{matrix}a;b;c>0\\a>b\end{matrix}\right.\)
\(B=\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}-1+2}{2^{10}-3+2}=\frac{2^{10}+1}{2^{10}-1}\)
\(\Rightarrow B>A\)
c) E=2010-1+2/2010-1=1+2/210-1
F=2010-3+2/2010-3=1+2/2010-3
Vì 2/210-1>2/2010-3 nên E>F
a) Ta có: 10A=1020+10/1020+1
=1020+1/1020+1-9/1020+1
=1-9/1020+1
10B=1021+10/1021+1
=1021+1/1021+1-9/1021+1
=1-9/1021+1
Vì 9/1020+1>9/1021+1 nên A>B
(Còn nữa)
Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
Bài 1:
a)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
b)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008\cdot\left(1-\frac{1}{2017}\right)\)
\(=1008\cdot\frac{2016}{2017}\)\(=\frac{290304}{31}\)Bài 2:
a)\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{2}{7}\)
b)\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{25.28}\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\cdot\frac{6}{28}\)
\(=\frac{15}{14}\)
Bài 3:
a)Đặt \(A=-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\)
\(=-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]\)
\(=-\left[10\cdot\frac{4}{55}\right]\)
\(=-\frac{8}{11}\).Thay vào ta có: \(x-\frac{8}{11}=\frac{2}{9}\)
\(\Leftrightarrow x=\frac{94}{99}\)
b)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)