a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)
Tính tỉ số \(\dfrac{A}{B}\) biết:
\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\) và
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\)
Lời giải:
\(299A=\frac{300-1}{1.300}+\frac{301-2}{2.301}+\frac{302-3}{3.302}+....+\frac{400-101}{101.400}\)
\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)
\(=(1+\frac{1}{2}+....+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(1)\)
Mặt khác:
$101B=\frac{102-1}{1.102}+\frac{103-2}{2.103}+...+\frac{400-299}{299.400}$
$=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}$
$=(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{299})-(\frac{1}{102}+\frac{1}{103}+....+\frac{1}{400})$
$=(1+\frac{1}{2}+...+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(2)$
Từ $(1);(2)\Rightarrow 299A=101B$
$\Rightarrow \frac{A}{B}=\frac{101}{299}$
1,
a,tính:\(\dfrac{\dfrac{7}{2012}+\dfrac{7}{9}-\dfrac{1}{4}}{\dfrac{5}{9}-\dfrac{1}{2012}-\dfrac{1}{2}}\)
b,so sánh:A=\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2010};B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{17}\)
M=\(\dfrac{\dfrac{7}{2012}+\dfrac{7}{9}-\dfrac{1}{4}}{\dfrac{5}{9}-\dfrac{3}{2012}-\dfrac{1}{2}}\) tính M
M=(\(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\)) : \(\dfrac{2012}{2013}\)
\(M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2012}{2013}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right):\dfrac{2012}{2013}\)
\(M=\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\right):\dfrac{2012}{2013}\)
\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right).\dfrac{2013}{2012}\)
\(M=0.\dfrac{2013}{2012}\)
\(M=0\)
Tính : S = \(1-\dfrac{1}{2}+\dfrac{1}{3}-\)\(\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2021}\)và
P = \(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+...+\dfrac{1}{2020}+\dfrac{1}{2021}\)
Tính : \(\left(S-P\right)^{2022}\)
S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)
= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)
Tìm \(\dfrac{a}{b}\)biết
A=\(\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+........+\dfrac{1}{101.400}\)
B=\(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+.........+\dfrac{1}{299.400}\)
Ta có :
\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+.............+\dfrac{1}{101.400}\)
\(299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...................+\dfrac{299}{101.400}\)
\(299A=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+..............+\dfrac{1}{101}-\dfrac{1}{400}\)
\(299A=\left(1+\dfrac{1}{2}+................+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+..............+\dfrac{1}{400}\right)=C\)
\(\Rightarrow A=\dfrac{C}{299}\)
Lại có :
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+................+\dfrac{1}{299.400}\)
\(101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...............+\dfrac{101}{299.400}\)
\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+..................+\dfrac{1}{299}-\dfrac{1}{400}\)
\(101B=\left(1+\dfrac{1}{2}+..............+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}-\dfrac{1}{103}+...............+\dfrac{1}{400}\right)=C\)
\(\Rightarrow B=\dfrac{C}{101}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{C}{101}:\dfrac{C}{299}=\dfrac{101}{299}\)
~ Chúc bn học tốt ~
1.tính M=\(\dfrac{\dfrac{7}{2012}+\dfrac{7}{9}-\dfrac{1}{4}}{\dfrac{5}{9}-\dfrac{3}{2012}-\dfrac{1}{2}}\)
\(M=\frac{\frac{7}{2012}+\frac{7}{9}-\frac{1}{4}}{\frac{5}{9}-\frac{3}{2012}-\frac{1}{2}}\)
\(M=\frac{\frac{63}{18108}-\frac{14084}{18108}-\frac{4527}{18108}}{\frac{10060}{18108}-\frac{27}{18108}-\frac{9054}{18108}}\)
\(M=\frac{-18548}{979}\)
Mình tin chắc đề bài sai , số to quá !!!
tính: \(C=\dfrac{\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+....+\dfrac{1}{101.400}}{\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+....+\dfrac{1}{299.400}}\)
