Cho M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+..........+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..........+\dfrac{1}{100}}\) ; N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-.........-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+......+\dfrac{1}{500}}\)
Tìm tỉ số phần trăm của M và N
3, Chứng tỏ :
\(B=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\)
a, \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2^2}\right)\left(1+\dfrac{1}{2^3}\right)\left(1+\dfrac{1}{2^4}\right)....\left(1+\dfrac{1}{2^{50}}\right)< 3\)
b, \(\dfrac{1}{2}-\dfrac{1}{2^2}+.........+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}< \dfrac{1}{3}\)
Chứng minh rằng
A= \(\dfrac{1}{2}\). \(\dfrac{3}{4}\).\(\dfrac{5}{6}\)..........\(\dfrac{99}{100}\)< \(\dfrac{1}{10}\)
B= 1+ \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+.......+ \(\dfrac{1}{64}\)>4
So sánh :
a) Chứng minh rằng : M = \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+.......+\dfrac{1}{100!} \)
Chứng minh rằng : M <1 .
b) Chứng minh rằng : N = \(\dfrac{9}{10!}+\dfrac{9}{11!}+\dfrac{9}{12!}+........+\dfrac{9}{1000!}\)
Chứng minh rằng : N < \(\dfrac{1}{9!}\)
cho M=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\)
N=\(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\)
chứng minh rằng: M<\(\dfrac{1}{10}\)
Em cần gấp câu trả lời cho bài toán này, mong đc mn giúp đỡ (nếu được xin trả lời trước 12h ngày 10/5 giúp em ạ). Cảm ơn mn.
Chứng minh:
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}và< \dfrac{1}{2}\)
\(a,\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}+\dfrac{-0,25.\dfrac{-2}{3}-75\%:\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)}{\left|-1\dfrac{1}{2}\right|.\left(\dfrac{-2}{3}-0,75:\dfrac{3}{-2}\right)}\)
cho tổng T= \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}\) +...+\(\dfrac{2016}{2^{2015}}+\dfrac{2017}{2^{2016}}\)
so sánh T với 3
