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\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{99.101}{100^2}\)
\(=\frac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4.....100.100}\)
\(=\frac{1.2.3.....99}{2.3.4.....100}.\frac{3.4.5.....101}{2.3.4.....100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
Ủng hộ mk nha,chúc bn học tốt!!!
\(C=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(C=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{99\cdot101}{100\cdot100}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)
\(C=\frac{1}{100}\cdot\frac{101}{2}\)
\(C=\frac{101}{200}\)
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x......x\frac{9999}{10000}\)
\(C=\frac{1.3}{2^2}x\frac{2.4}{3^2}x\frac{3.5}{4^2}x....x\frac{99.101}{100^2}\)
\(C=\frac{1.3.2.4.3.5.......99.101}{2^2.3^2.4^2.......100^2}\)
\(C=\frac{1.2.3.......99}{2.3.4....100}x\frac{3.4.5.....101}{2.3.4......100}\)
\(C=\frac{1}{100}.\frac{101}{2}=\frac{1.101}{100.2}=\frac{101}{200}\)
Ủng hộ mk nha!!!!
3/4.8/9.15/16.....9999/10000
=1.3/2^2.2.4/3^2.3.5/4^2....99.101/100^2
=1.3/2.2.2.4/3.3.3.5/4.4....99.101/100.100
=(1.2.3...99/2.3.4...100).(3.4.5...101/2.3.4...100)
=1/100.101/2=101/200
tách trên tử thành 2 stn hon kem nhau 2 dvi o duoi mau la binh phuong cua mot so
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(C=\frac{3}{4}x\frac{4x2}{3x3}x\frac{3x5}{2x8}x...x\frac{99x101}{100x100}\)
\(C=...\) ( Tự làm tiếp )
\(E=1\frac{1}{3}x1\frac{1}{8}x1\frac{1}{15}x1\frac{1}{24}x...x1\frac{1}{99}\)
\(E=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x...x\frac{100}{99}\)
\(E=....\)( tương tự câu C )
\(M=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot....\cdot\frac{99\cdot101}{100\cdot100}=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot99\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\frac{1\cdot101}{2\cdot100}=\frac{101}{200}\)Vậy M = \(\frac{101}{200}\)
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(M=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}=\frac{3.8.15...9999}{4.9.16...10000}=\frac{1.3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}=\frac{\left(1.2.3...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(\frac{1.101}{100.2}=\frac{101}{200}\)
Ta có : \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{15}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)< 99\)
\(\Rightarrow\)S<99 (1)
Đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}< 1\)
\(\Rightarrow\)S>99-1=98 (2)
Từ (1) và (2)
\(\Rightarrow\)98<S<99
\(\Rightarrow\)S\(\notin\)N
Vậy S\(\notin\)N.
\(B=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right)...\left(99.101\right)}{2^2.3^2.4^2.5^2...100^2}=\frac{\left(1.2.3.4...99\right).\left(3.4.5.6...101\right)}{\left(2.3.4.5...100\right)\left(2.3.4.5...100\right)}=\frac{1.101}{100.2}=\frac{101}{200}\)
B = \(\frac{1.3}{2^2}.\frac{2.4}{3^2}\frac{3.5}{4^2}\frac{4.6}{5^2}...\frac{99.101}{100^2}=\frac{1.3.2.4.3.5.4.6...99.101}{2.2.3.3.4.4.5.5...100.100}\)
=\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
Vật B = \(\frac{101}{200}\)
đúng cái đi
B=$\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}=?$34 .89 .1516 .2425 ...999910000 =?