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\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}\)
\(\Rightarrow C=\frac{6}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)
\(\Rightarrow C=3.\left(\frac{1}{3}-\frac{1}{47}\right)\)
\(\Rightarrow C=3.\frac{44}{141}\)
\(\Rightarrow C=\frac{44}{47}\)
\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{45.47}\right)=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\\ \)
\(=3.\left(\frac{1}{3}-\frac{1}{47}\right)=\frac{3.44}{141}=\frac{44}{47}\)
khỏi ghi lại đề nha
A=1-1/2+1/2-1/3+1/3-1/4+......+1/49-1/50
A=1-1/50
A=49/50
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\frac{1}{2004}=\frac{2003}{2004}\)
b) Đặt A=\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2003\cdot2005}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(2A=1-\frac{1}{2005}\)
\(2A=\frac{2004}{2005}\)
\(A=\frac{2004}{2005}:2=\frac{2004}{2005}\cdot\frac{1}{2}=\frac{1002}{2005}\)
a)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(=\frac{1}{1}-\frac{1}{2004}\)
\(\Rightarrow=\frac{2003}{2004}\)
b)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003+2005}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(=\frac{1}{1}-\frac{1}{2005}\)
\(\Rightarrow=\frac{2004}{2005}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{99.100}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}+\frac{1}{100}=\frac{49}{100}\)
Đề \(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\) chứ bn!
Bài làm
ta có: 1.98 +2.97 + 3.96 + 98.1
= 1 + (1+2) + ( 1+2+3) +...+ ( 1+2+3+...+ 98)
\(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}\)
\(=\frac{1.2+2.3+3.4+...+98.99}{2}\)
\(\Rightarrow A=\frac{1.98+2.99+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{\left(1.2+2.3+3.4+...+98.99\right).\frac{1}{2}}{1.2+2.3+3.4+...+98.99}\)
\(A=\frac{1}{2}\left(đpcm\right)\)
\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(b,\frac{x}{y}=\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{3}=\frac{y}{5}\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\)
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{18}{8}=\frac{9}{4}\)
\(\Rightarrow\frac{x}{3}=\frac{9}{4}\Rightarrow x=\frac{27}{4}\)
\(\frac{y}{5}=\frac{9}{4}\Rightarrow y=\frac{45}{4}\)
\(\left(1+\frac{1}{4}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right).\left(1+\frac{1}{24}\right)...\left(1+\frac{1}{9999}\right)\)
\(=\frac{5}{4}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{10000}{9999}=\frac{5.9.16.25...10000}{4.8.15.24...9999}=\frac{5.3^2.4^2.5^2...100^2}{4.2.4.3.5.4.6...99.101}\)
\(=\frac{5.3.4.5...100.3.4.5...100}{4.2.3.4...99.4.5.6...101}=\frac{5.100.3}{4.2.101}=\frac{5.25.3}{2.101}=\frac{375}{202}.\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{200}-\frac{1}{201}\)
\(=\frac{1}{2}-\frac{1}{201}\)
\(=\frac{201}{402}-\frac{2}{402}\)
\(=\frac{199}{402}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{200}-\frac{1}{201}\)
\(=\frac{1}{2}-\frac{1}{201}\)
\(=\frac{199}{402}\)