1<1/3+1/8+1/15+1/24+....+1/360>

KO BIẾT ĐÚNG HAY KO NHÉ BẠN 

\(1\frac{1}{3}\)là hỗn số mà

\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}...1\frac{1}{360}\)

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{361}{360}\)

\(=\frac{2^2}{3}.\frac{3^2}{8}.\frac{4^2}{15}.\frac{5^2}{24}...\frac{19^2}{360}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}...\frac{19.19}{18.20}\)

\(=\left(\frac{2.3.4.5...19}{1.2.3.4...18}\right).\left(\frac{2.3.4.5...19}{3.4.5.6...20}\right)\)

\(=19.\frac{1}{10}\)

\(=\frac{19}{10}\)

\(A=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.....1\frac{1}{360}\)

\(A=1+\left(\frac{1}{3}.\frac{1}{8}.\frac{1}{15}.\frac{1}{24}.....\frac{1}{360}\right)\)

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 \(\frac{2^2}{3}.\frac{3^2}{8}......\frac{99^2}{9800}=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}=\frac{2.2.3.3.....99.99}{1.3.2.4....98.100}\)

\(=\frac{\left(2.3.4...99\right).\left(2.3.4....99\right)}{\left(1.2.3....98\right).\left(3.4.5...100\right)}=\frac{99.2}{1.100}=\frac{198}{100}=\frac{99}{50}\)

giúp với !

\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}.\frac{36}{35}......\frac{9801}{9800}=\frac{\left(2.3.4.5....99\right)^2}{1.3.2.4.3.5.4.6.....98.100}=\frac{2.3.4.5...99}{1.2.3.4.....98}.\frac{2.3.4.5....99}{3.4.5.6......100}=\frac{99}{1}.\frac{2}{100}=\frac{99}{50}\)

A=4/3.9/8.16/15.25/24. ... .9801/9800

Tính được chưa

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\(\frac{2012}{2011}\)

=\(\frac{2012}{9}\)

\(1\frac{1}{9}.1\frac{1}{10}.1\frac{1}{11}...1\frac{1}{2011}\)

= \(\frac{10}{9}.\frac{11}{10}.\frac{12}{11}...\frac{2012}{2011}\)

= \(\frac{2012}{9}\)

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

\(1\frac{1}{38}.1\frac{1}{39}.1\frac{1}{40}...1\frac{1}{2013}=\frac{39}{38}.\frac{40}{39}.\frac{41}{40}...\frac{2014}{2013}=\frac{2014}{38}=53\)

Do thấy họ của bạn giống họ của tui nên giúp

#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

3)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{49}{50}\)

\(=\frac{49}{50}\)

⇒ A < 1 (1)

\(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)

\(\Rightarrow B>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{10}+\frac{90}{100}=1\)

⇒ B > 1 (2)

từ (1) và (2) ⇒ A<1<B

vậy A < B

    \(\frac{15}{39}.\left(7\frac{4}{5}.1\frac{2}{3}+8\frac{1}{3}.7,8\right)\)

\(=\frac{15}{39}.\left(7,8.1\frac{2}{3}+8\frac{1}{3}.7,8\right)\)

\(=\frac{5}{13}.\left\{7,8.\left(1\frac{2}{3}+8\frac{1}{3}\right)\right\}\)

\(=\frac{5}{13}.\left(7,8.10\right)\)

\(=\frac{5}{13}.78\)

\(=30\)