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1
27 tháng 4 2017
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2009}{2011}\)
Đặt tổng vế trái là A
Ta có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\)
\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)\div2}\right)\)
\(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\)
\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)
\(\frac{1}{2}A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)
\(A=\left(\frac{1}{2}+\frac{1}{x+1}\right):\frac{1}{2}\)
\(A=1+\frac{1}{\left(x+1\right)\div2}\)
\(\Rightarrow1+\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}-1=\frac{2009}{2011}-\frac{2011}{2011}=-\frac{2}{2011}\)
\(\Rightarrow-\frac{2}{-\left(x+1\right)}=-\frac{2}{2011}\)
\(\Rightarrow-\left(x+1\right)=2011\)
\(\Rightarrow x+1=-2011\)
\(\Rightarrow x=-2011-1=-2012\)
19 tháng 1 2022
a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)
b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)
ND
0
13 tháng 9 2018
\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...+\frac{1}{16384}\)
\(A=\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{14}}\)
\(2^2A=1+\frac{1}{2^2}+...+\frac{1}{2^{12}}\)
\(4A-A=\left(1+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{14}}\right)\)
\(3A=1-\frac{1}{2^{14}}\)
\(A=\frac{1-\frac{1}{2^{14}}}{3}\)
