S = 1/3 + 1/32 + ... + 1/39
=> 3S = 1 + 1/3 + ... + 1/38
=> 3S - S = 1 + 1/3 + ... + 1/38 - 1/3 - 1/32 + ... - 1/39
=>2S = 1 - 1/39 = (39 - 1)/39
=> S = (39 - 1)/(2.39)
Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiều
a)\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}+\frac{1}{3}\)
b) \(\frac{\frac{1}{3}-\frac{1}{5}-\frac{1}{7}}{\frac{2}{3}-0,4-\frac{2}{7}}+\frac{\frac{3}{8}-\frac{3}{16}-\frac{3}{32}+\frac{3}{64}}{\frac{1}{4}-\frac{1}{8}-\frac{1}{16}+\frac{1}{32}}\)
c) \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
Cho S =\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+\(\frac{1}{3^4}\)+\(\frac{1}{3^5}\)+\(\frac{1}{3^6}\)+\(\frac{1}{3^7}\)+\(\frac{1}{3^8}\)+\(\frac{1}{3^9}\)
So sánh S và 1
\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}+\frac{1}{3^9}=?\)
Tính
a) \(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{1}{-8}+\frac{1}{9}+\frac{1}{8}+\frac{1}{-7}+\frac{-1}{6}+\frac{-1}{5}\)
b) (-11).36-64.11
c) \(\frac{\frac{1}{3}+\frac{1}{7}+\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}+\frac{2}{13}}.\frac{\frac{3}{4}+\frac{3}{16}+\frac{3}{64}+\frac{3}{256}}{1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}}+\frac{3}{8}\)
\(S=\frac{1}{2}:\frac{3}{2}:\frac{4}{3}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}:\frac{9}{8}:\frac{10}{9}\)
Bài 1:
a, Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) .Chứng minh rằng \(\frac{2}{5}< S< \frac{8}{9}\)
b, Tìm x thuộc z để phân số \(\frac{x^2-5x-1}{x+2}\)có giá trị là số nguyên
c, Chứng minh rằng \(\left(\frac{7}{65}+1\right)\left(\frac{7}{84}+1\right)\left(\frac{7}{105}+1\right)\left(\frac{7}{124}+1\right)...\left(\frac{7}{153+1}\right)\left(\frac{7}{560}+1\right)< 2\)
d, Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Tính : \(K=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right).3^5+\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}.3^9\right)+...+\left(\frac{1}{3^{97}}+\frac{1}{3^{98}}+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right).3^{101}\)
A = \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}-1\right)\times\left(1-\frac{8}{1}-\frac{4}{1}-\frac{2}{1}\right)\)
B = \(\frac{\frac{3}{1}-\frac{6}{3}-\frac{9}{6}-\frac{369}{1}}{\frac{1}{3}+\frac{3}{6}+\frac{6}{9}-\frac{1}{963}}\)
C = \(\frac{1}{1}-\frac{1}{2}+\frac{3}{1}-\frac{1}{4}+\frac{5}{1}-\frac{1}{6}+\frac{7}{1}-\frac{1}{8}+\frac{9}{1}-\frac{1}{10}\)
so sánh các số trên ( A , B , C )
