tính:
\(1\frac{1}{38}\cdot1\frac{1}{39}\cdot1\frac{1}{40}\cdot\)...(cho đến)...\(\cdot1\frac{1}{2013}\)
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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
1 1/9 x 1 1/10 x 1 1/11 x ... x 1 1/2011
=10/9 x 11/10 x 12/11 x ... x 2012/2011
khử
còn 2012/9
=\(\frac{10}{9}\)x\(\frac{11}{10}\)x\(\frac{12}{11}\)x.........x\(\frac{2012}{2011}\)
=\(\frac{2012}{9}\)
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
\(\frac{15}{39}.\left(7\frac{4}{5}.1\frac{2}{3}+8\frac{1}{3}.7,8\right)\)
\(=\frac{15}{39}.\left(7,8.1\frac{2}{3}+8\frac{1}{3}.7,8\right)\)
\(=\frac{5}{13}.\left\{7,8.\left(1\frac{2}{3}+8\frac{1}{3}\right)\right\}\)
\(=\frac{5}{13}.\left(7,8.10\right)\)
\(=\frac{5}{13}.78\)
\(=30\)
Bg
\(1\frac{1}{38}.1\frac{1}{39}.1\frac{1}{40}\)...\(.1\frac{1}{2013}\)
= \(\frac{39}{38}.\frac{40}{39}.\frac{41}{40}.\)...\(.\frac{2014}{2013}\)
= \(\frac{39.40.41.....2014}{38.39.40......2013}\)(39 trên, 39 dưới, 40 trên, 40 dưới,... 2013 trên, 2013 dưới, chịt tiêu hết)
= \(\frac{2014}{38}\)
= 53
cảm ơn bẹn nhe