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\(A=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.....1\frac{1}{360}\)
\(A=1+\left(\frac{1}{3}.\frac{1}{8}.\frac{1}{15}.\frac{1}{24}.....\frac{1}{360}\right)\)
Nếu đúng thì tk nha
Ta có \(1\frac{1}{3}=\frac{2^2}{3};1\frac{1}{8}=\frac{3^2}{8};.....\)
Nên thừa số thứ 98 là : \(1\frac{1}{9800}=\frac{99^2}{9800}\)
Ta có \(\frac{2^2}{3}.\frac{3^2}{8}......\frac{99^2}{9800}=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}=\frac{2.2.3.3.....99.99}{1.3.2.4....98.100}\)
\(=\frac{\left(2.3.4...99\right).\left(2.3.4....99\right)}{\left(1.2.3....98\right).\left(3.4.5...100\right)}=\frac{99.2}{1.100}=\frac{198}{100}=\frac{99}{50}\)
#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{9800}=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{9801}{9800}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{99.99}{98.100}\)
\(=\frac{2.3.4...99}{1.2.3....98}.\frac{2.3.4...99}{3.4.5...100}\)
\(=99.\frac{2}{100}=99.\frac{1}{50}=\frac{99}{50}\)
a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}=\frac{31}{32}\)
b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}.\frac{36}{35}......\frac{9801}{9800}=\frac{\left(2.3.4.5....99\right)^2}{1.3.2.4.3.5.4.6.....98.100}=\frac{2.3.4.5...99}{1.2.3.4.....98}.\frac{2.3.4.5....99}{3.4.5.6......100}=\frac{99}{1}.\frac{2}{100}=\frac{99}{50}\)
A=4/3.9/8.16/15.25/24. ... .9801/9800
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