Tinh giai tri cac bieu thuc sau
F = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{190}\)
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1/2D=1/2(1/6+1/10+......+1/45)
1/2D=1/12+1/20+1/30+.....+1/90
1/2D=1/3.4+1/4.5+1/5.6+......+1/9.10
1/2D=1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10
1/2D=1/3-1/10
1/2D=7/30
D=7/30:1/2
D=7/15
Ta có:\(D=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(=\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)
\(=2.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{10}\right)=2.\frac{7}{30}=\frac{7}{15}\)
Vậy \(D=\frac{7}{15}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
Thêm đk \(a,b,c\ne0\)
Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)
\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)
\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)
\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)
\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)
\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
\(\frac{F}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)
\(\frac{F}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(\frac{F}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.5}+\frac{5-4}{4.5}+...+\frac{20-19}{19.20}\)
\(\frac{F}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(\frac{F}{2}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow F=\frac{18}{20}=\frac{9}{10}\)