B=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}=?\)
Cho \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{9999}{10000}\). Chứng minh rằng : S\(\notin\)N.
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.
Tính giá trị của A biết:
1/ \(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}\)
2/ \(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{3599}{3600}\)
mọi nguời cứ tính đi đuợc câu nào tôi cũng tick cho
Chứng tỏ rằng \(\frac{3}{4}\)+\(\frac{8}{9}\)+\(\frac{15}{16}\)+\(\frac{24}{25}\)+....+\(\frac{9999}{10000}\)không phải là số nguyên
Đặt A =\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)
Đặt B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}\)
Khi đó A = \(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)< 99-\frac{99}{202}\approx98,5\)
=> A < 98,5 (1)
Lại có B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Khi đó A =\(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)>99-\frac{99}{100}=98,01\)
=> A > 98,01 (2)
Từ (1)(2) => 98,01 < A < 98,5
=> A không là số nguyên
Tính x=\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(x=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(x=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(x=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)
\(x=\frac{1}{100}.\frac{101}{2}\)
\(x=\frac{101}{200}\)
\(X=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(X=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)
\(X=\frac{1}{100}.\frac{101}{2}\)
\(X=\frac{101}{200}\)
Study well
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\frac{3.8.15....9999}{4.9.16....10000}=?\)
Tính \(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(M=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot....\cdot\frac{99\cdot101}{100\cdot100}=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot99\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\frac{1\cdot101}{2\cdot100}=\frac{101}{200}\)Vậy M = \(\frac{101}{200}\)
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(M=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x.......x\frac{9999}{10000}\)
\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{99.101}{100^2}\)
\(=\frac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4.....100.100}\)
\(=\frac{1.2.3.....99}{2.3.4.....100}.\frac{3.4.5.....101}{2.3.4.....100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
Ủng hộ mk nha,chúc bn học tốt!!!
bn ơi mjk chưa học số có mũ bn giúp mjk nốt nha
C=\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x.......x\frac{9999}{10000}\)
\(C=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{10000}\)
\(C=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{99\cdot101}{100\cdot100}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot4\cdot...\cdot100}\)
\(C=\frac{1}{100}\cdot\frac{101}{2}\)
\(C=\frac{101}{200}\)
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x......x\frac{9999}{10000}\)
\(C=\frac{1.3}{2^2}x\frac{2.4}{3^2}x\frac{3.5}{4^2}x....x\frac{99.101}{100^2}\)
\(C=\frac{1.3.2.4.3.5.......99.101}{2^2.3^2.4^2.......100^2}\)
\(C=\frac{1.2.3.......99}{2.3.4....100}x\frac{3.4.5.....101}{2.3.4......100}\)
\(C=\frac{1}{100}.\frac{101}{2}=\frac{1.101}{100.2}=\frac{101}{200}\)
Ủng hộ mk nha!!!!
B=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{9999}{10000}\)
Tính B