Câu hỏi của lương hiếu - Toán lớp 6 - Học toán với OnlineMath
Làm như link trên nhưng bỏ hạng tử \(\frac{1}{99.100}\)đi
Bước cuối: \(1-\frac{1}{99}=\frac{98}{99}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(\Rightarrow C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow C=1-\frac{1}{99}\)
\(\Rightarrow C=\frac{98}{99}\)
~Study well~
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=1-1/2 +1/2-1/3 +1/3-1/4 +....+1/97.98+1/98+1/99
=1-1/99
=98/99
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(\Rightarrow C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow C=1-\frac{1}{99}\)
\(\Rightarrow C=\frac{98}{99}\)
rất vui vì giúp đc bạn !!!
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}\)
\(=1-\frac{1}{98}\)
\(=\frac{98}{98}-\frac{1}{98}\)
\(=\frac{97}{98}\)
\(C=\frac{1}{1.2} +\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{1}-\frac{1}{99}\)
\(=\frac{98}{99}\)