\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\) \(C=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
\(B=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\) \(D=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}< \frac{1}{100}\)
Mọi người giúp mik nhé, mik đang ôn thi nên cần gấp!
Bài 1 : Chứng minh
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b) \(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{9999}{10000}< \frac{1}{100}\)
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}< 2^{ }\)
b)\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c)\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
a, A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< 1\)
b, B=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}< 2\)
c, C=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}< 6\)
d, D=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{9999}{10000}< \frac{1}{100}\)
1)Chứng minh các phân số sau là các phân số tối giản:
a)\(A=\frac{12n+1}{30n+2}\)
b)\(B=\frac{14n+17}{21n+25}\)
2)Chứng minh rằng:
a)\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b)\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c)\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
Bài 1: Tìm x, biết:
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+.....+\frac{1}{49.50}x=1\)
Bài 2: Chứng minh rằng:
\(a)A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}< 2\)
\(b)B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{63}< 6\)
\(c)C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.......\frac{9999}{10000}< \frac{1}{100}\)
Bài 3: Tính tổng:
\(S=\frac{1+2+2^2+2^3+.....+2^{2008}}{1-2^{2009}}\)
Chứng minh:
a, A = 1 + \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ ... + \(\frac{1}{100^2}\)< 2
b, B = 1 + \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ .... + \(\frac{1}{63}\)< 6
c, C = \(\frac{1}{2}\). \(\frac{3}{4}\). \(\frac{5}{6}\). \(\frac{7}{8}\).... \(\frac{9999}{10000}\)< \(\frac{1}{100}\)
GIÚP MÌNH VỚI NHÉ, MÌNH ĐANG CẦN GẤP LẮM
chứng minh :
a,A=1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)< 2
b,B=1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+...+\(\frac{1}{63}\)< 6
c,C=\(\frac{1}{2}\).\(\frac{3}{4}\).\(\frac{5}{6}\).\(\frac{7}{8}\)...\(\frac{9999}{10000}\)< \(\frac{1}{100}\)
cm
a,A=1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+....+\(\frac{1}{100^2}\)<2
b,B=1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+....+\(\frac{1}{63}\)<6
c,C=\(\frac{1}{2}\).\(\frac{3}{4}\).\(\frac{5}{6}\).\(\frac{7}{8}\)...\(\frac{9999}{10000}\)<\(\frac{1}{100}\)
3.Tính hợp lí:
a,\(\frac{17}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}\)
b,\(\left(\frac{11}{4}.\frac{5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)
c,\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
4.Tìm tích:
a,\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right).....\left(\frac{1}{99}+1\right)\)
b,\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{100}-1\right)\)
c,\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}....\frac{899}{30^2}\)
AI CÒN THỨC THÌ GIÚP MIK VS,MIK ĐANG CẦN GẤP
