a, tổng các tử và mẫu mỗi phân sô trên đều bằng 200
b, \(A=\dfrac{1}{199}+\dfrac{2}{198}+\dfrac{3}{197}+...+\dfrac{198}{2}+\dfrac{199}{1}\)
\(A=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)
\(A=200\left(\dfrac{1}{199}+\dfrac{1}{198}+...+\dfrac{1}{2}+\dfrac{1}{200}\right)\)(đpcm)
tính
A=\(\dfrac{\dfrac{1}{199}+\dfrac{1}{198}+.....+\dfrac{1}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{200}}\)
Tính \(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)
S=\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{199}{200}\)
CMR S2<\(\dfrac{1}{200}\)
\(\dfrac{1}{200}< \dfrac{1}{100^2}+\dfrac{1}{101^2}+...+\dfrac{1}{199^2}< \dfrac{1}{99}\)
Chứng minh rằng:
\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{7}{12}\)
Giúp mk nhé!
so sanh
M=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)va \(\dfrac{1}{2}\)
B=\(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}va\dfrac{9}{10}\)
C=\(\dfrac{10}{17}+\dfrac{8}{15}+\dfrac{11}{16}va2\)
chứng minh \(1^2+2^2+...+n^2=\dfrac{1}{3}n\left(n+\dfrac{1}{2}\right)\left(n+1\right)=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n\)
Chứng minh rằng :
a) \(\dfrac{1.3.5.....39}{21.22.23.....40}=\dfrac{1}{2^{20}}\)
b) \(\dfrac{1.3.5....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\dfrac{1}{2^n}\) với \(n\in\) N*
1, so sánh A;B biết: A=\(\left(\dfrac{\left(3\cdot\dfrac{2}{15}+\dfrac{1}{5}\right):2\cdot\dfrac{1}{2}}{\left(5\cdot\dfrac{3}{7}-2\cdot\dfrac{1}{4}\right):\dfrac{443}{56}}\right);B=\dfrac{1,2:\left(1\cdot\dfrac{1}{5}.1\cdot\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
