\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Nhầm tưởng tính tích :v
Ta có :
\(B=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}+\frac{1}{100}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=50.\frac{1}{51}=\frac{50}{51}< \frac{99}{100}\)
\(\Leftrightarrow A>B\)
~ Rim Ceil ~:Chuyên Quốc Học ở đâu thì ko biết nhưng bài như thế này mak làm sai~
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{100}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+....+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow A-B=0\)