Violympic toán 7
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30 tháng 9 2023 lúc 15:00
tính tỉ số \(\dfrac{A}{B}\) biết A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{101.102}\) và B=\(\dfrac{1}{52.102}\)+\(\dfrac{1}{53.101}\)+...+\(\dfrac{1}{102.52}\)+\(\dfrac{2}{77.154}\)
CMR : \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
Tính \(P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}}\)
chứng minh rằng
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
CMR:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
A=\(\dfrac{1}{1.2^2}+\dfrac{1}{2.3^2}+\dfrac{1}{3.4^2}+...+\dfrac{1}{49.50^2}\)
B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
CM A<B
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...-\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\)
Ghi cách giải lun nha
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
CMR
\(\dfrac{7}{12}< A< \dfrac{5}{6}\)
1. Chứng minh rằng:
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100}< 1\)
2. Chứng minh rằng:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100-1}{100!}< 2\)