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\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
\(S=\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}\right)-\left(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}\right).\)
Đặt A là biểu thức trong ngoặc đơn thứ nhất bà B là biểu thức trong ngoặc đơn thứ 2
\(2A=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
\(A=\frac{8}{9}:2=\frac{4}{9}\)
\(2B=\frac{4-2}{2x4}+\frac{6-4}{4x6}+\frac{8-6}{6x8}=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}\)
\(2B=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\Rightarrow B=\frac{3}{8}:2=\frac{3}{16}\)
\(S=A-B=\frac{4}{9}-\frac{3}{16}\)
\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+..+\dfrac{1}{97.99}+\dfrac{1}{98.100}-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{99.100}\right)\right]-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{98}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[1-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)
\(=\dfrac{1}{2}\left[\dfrac{98}{99}+\dfrac{49}{100}\right]-\dfrac{49}{99}=\dfrac{14651}{19800}-\dfrac{49}{99}=\dfrac{49}{200}\)
\(\dfrac{1}{1x3}+\dfrac{1}{2x4}+...+\dfrac{1}{98x100}+\dfrac{1}{97x99}-\dfrac{49}{99}=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{100}-\dfrac{49}{99}=1-\dfrac{1}{100}-\dfrac{49}{99}\)
=\(\dfrac{4901}{9900}\)
\(A=\left(1-\frac{1}{15}\right).\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right)......\left(1-\frac{1}{1275}\right)\)
\(S=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+\dfrac{1}{5.7}\)
\(S=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}\)
\(S=1+\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{31}{21}\)
Chúc bạn học tốt!!!
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Ta có:
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{98.100}\)
\(\Rightarrow2A=\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{98.100}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)
Vậy giá trị của biểu thức là \(\frac{49}{200}\)