i love you lê nguyễn minh thiên ^_^
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
Ta có công thức:\(\frac{a}{b.c}=\frac{a}{c-b}\left[\frac{1}{b}-\frac{1}{c}\right]\)
Vậy \(B=1\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right]\)
\(B=1\left[1-\frac{1}{100}\right]\)
\(=\frac{99}{100}\)
Nếu chưa có ai trả lời hoặc cậu đã có rồi cũng không sao
Chúc bạn học tốt!
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9702}+\frac{1}{9900}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
~Học tốt~
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{9702}+\frac{1}{9900}\)
\(B=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
\(#Jen\)
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