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\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{10.11}{2}}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)=2\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
a,11/5x3/2-3/1/2x3/5
=11/2x3/5-7/2-3/5
=3/5x(11/2-7/2)
=3/5x2
=6/5
b,(1/2-1/3+1/4-1/5):(1/4-1/6)
=(1/6+1/20):1/12
=13/60:1/2
=13/30
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4+5}\) = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{15}\)= \(\frac{10}{30}+\frac{5}{30}+\frac{2}{30}\)= \(\frac{10+5+2}{30}\)= \(\frac{17}{30}\)
Trong đề có 4 đáp án là a)\(\frac{1}{6}\)
b)\(\frac{5}{6}\)
c) \(\frac{1}{3}\)
d)\(\frac{2}{3}\)
\(=\frac{2008+\left(1+\frac{2007}{2}\right)+...+\left(1+\frac{1}{2008}\right)}{\frac{1}{2}+...\frac{1}{2009}}-2007\)
\(=\frac{1+\frac{2009}{2}+...\frac{2009}{2008}}{\frac{1}{2}+...+\frac{1}{2009}}\)
\(=\frac{\frac{2009}{2009}+\frac{2009}{2}+...+\frac{2009}{2008}}{\frac{1}{2}+...+\frac{1}{2009}}\)
\(=\frac{2009\left(\frac{1}{2}+...+\frac{1}{2009}\right)}{\frac{1}{2}+...+\frac{1}{2009}}=2009\)
\(M=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}\)
\(M=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+\frac{1}{\left(1+5\right).5:2}\)
\(M=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}\)
\(M=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(M=2.\frac{1}{2}-2.\frac{1}{6}\)
\(M=1-\frac{1}{3}=\frac{2}{3}\)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+10}=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{10.11}{2}}\)(dựa vào công thức:
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)(với n là số nguyên dương). Nên biểu thức sẽ bằng:
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}=2\left(\frac{1}{2.3}+...+\frac{1}{10.11}\right)=2\left(\frac{3-2}{2.3}+....+\frac{11-10}{10.11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{11}\right)=2\left(\frac{1}{2}-\frac{1}{11}\right)=1-\frac{2}{11}=\frac{9}{11}\).
Vậy đáp án là: \(\frac{9}{11}\).