A<50/100+50/100+50/100+50/100=4.50/100=2
=>A<2
A>4.50/150=4/3+1+1/3>1
=>dccm
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Các câu hỏi tương tự
Cho A= \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\). CMR 50<A<100
cho A= \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + .............+ \(\dfrac{1}{50}\)
B= \(\dfrac{49}{1}\) + \(\dfrac{48}{2}\) + .....+ \(\dfrac{1}{49}\) = 50 * B
CMR A và B ko là số tự nhiên
1 CM
a, \(\left(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{2n-1}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2n}\right)=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{2n}\)( n∈Z)
b, \(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}=\dfrac{99}{50}-\dfrac{97}{49}+...+\dfrac{7}{4}-\dfrac{5}{3}+\dfrac{3}{2}\)
Bài 1: tính
Cho A dfrac{1}{31}+dfrac{1}{32}+dfrac{1}{33}+........+dfrac{1}{60}dfrac{7}{12}
Bdfrac{1}{3^2}+dfrac{1}{3^2}+dfrac{1}{5^2}+.....+dfrac{1}{50^2}
CMR B dfrac{1}{4}; B dfrac{4}{9}
C dfrac{1}{2}.dfrac{3}{4}.dfrac{5}{6}.dfrac{7}{8}...........dfrac{79}{80}dfrac{1}{9}
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Bài 1: tính
Cho A= \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+........+\dfrac{1}{60}>\dfrac{7}{12}\)
B=\(\dfrac{1}{3^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{50^2}\)
CMR B > \(\dfrac{1}{4}\); B < \(\dfrac{4}{9}\)
C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}...........\dfrac{79}{80}\)<\(\dfrac{1}{9}\)
Cho S=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)
và P=\(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
TÍNH S/P
Cho S = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\)và P = \(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\). Tính \(\dfrac{S}{P}\)
-45 : 2\(\dfrac{4}{7}\) - 50% + 1,25
Câu 3:
a, Chứng minh rằng nếu:
(overline{ab}+overline{cd}+overline{eg}) ⋮ 11 thì overline{abcdeg} ⋮ 11
b, Cho E 92-dfrac{1}{9}-dfrac{2}{10}-dfrac{3}{11}-...-dfrac{92}{100}; F dfrac{1}{45}+dfrac{1}{50}+dfrac{1}{55}+...+dfrac{1}{500}Tính dfrac{E}{F}
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Câu 3:
a, Chứng minh rằng nếu:
(\(\overline{ab}\)+\(\overline{cd}\)+\(\overline{eg}\)) ⋮ 11 thì \(\overline{abcdeg}\) ⋮ 11
b, Cho E = 92-\(\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}\); F= \(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\)Tính \(\dfrac{E}{F}\)
\(\dfrac{2}{3}\)✖x +50%+x=\(\dfrac{5}{4}\)