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a; A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 2
1 = 1 = 1
1/2^2 < 1/1.2 = 1/1 - 1/2
1/3^2 < 1/2.3 = 1/2 - 1/3
.......................
1/100^2 < 1/99.100 = 1/99 - 1/100
Cộng vế với vế ta có:
A = 1 + 1 - 1/100
A = 2 - 1/100 < 2 (đpcm)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{3.8.15...9999}{4.9.16...10000}=\frac{1.3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}=\frac{\left(1.2.3...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(\frac{1.101}{100.2}=\frac{101}{200}\)
1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)
a) = 3/3 x ( -24/54 +45/54 ) : 7/12
= 1 x 21/54 x 12/7
= 18/27
( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )
S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)
S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)
S = \(\frac12\) - \(\frac19\)
S = \(\frac{9}{18}-\frac{2}{18}\)
S = \(\frac{7}{18}\)
Ta có: \(\frac{3}{4}=1-\frac{1}{4}=1-\frac{1}{2^2}\); \(\frac{8}{9}=1-\frac{1}{9}=1-\frac{1}{3^2}\)
\(\frac{15}{16}=1-\frac{1}{16}=1-\frac{1}{4^2}\); ...; \(\frac{9999}{10000}=1-\frac{1}{10000}=1-\frac{1}{100^2}\)
=> \(C=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
=> \(C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)=99-B\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=> \(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
=> A > 99-1 = 98
=> B > 98
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a)\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\)\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}=\frac{101}{2.100}=\frac{101}{200}\)
b)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{3599}{3600}=\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.6}.....\frac{59.61}{60.60}=\frac{2.61}{60}=\frac{61}{30}\)