(1 -1/4)(1-1/9)(1-1/16)...(1-1/n2)
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ML
1
NN
0
NN
1
NN
1
26 tháng 7 2023
`= 3/4 . 8/9 . 15/16. ... . 143/144`
`= (1.2.3...12)/(2.3.4....12) . (3.4.....12.13)/(2.3....11.12)`
`= 1/12 . 13/2 = 13/24`
26 tháng 7 2023
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)*...*(1-1/12)(1+1/12)
=1/2*2/3*...*11/12*3/2*4/3*...*13/12
=1/12*13/2=13/24
11 tháng 7 2021
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
NT
1
BT
12 tháng 4 2018
Ta có: \(1-\frac{1}{4}=\frac{3}{4}=\frac{1}{2}.\frac{3}{2}\); \(1-\frac{1}{9}=\frac{8}{9}=\frac{2}{3}.\frac{4}{3}\); \(1-\frac{1}{16}=\frac{15}{16}=\frac{3}{4}.\frac{5}{4}\);
...; \(1-\frac{1}{10000}=\frac{9999}{10000}=\frac{99}{100}.\frac{101}{100}\)
=> \(A=\frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}.\frac{3}{4}.\frac{5}{4}....\frac{99}{100}.\frac{101}{100}\). Nhận thấy; Tích của 2 số liền kề thì bằng 1
=> \(A=\frac{1}{2}.\frac{101}{100}=\frac{101}{200}\)
Đáp số: \(A=\frac{101}{200}\)
CM
3 tháng 3 2017
a) x 2 – x – 12 b) x 3 – 64.
c) m 3 n 3 – m 2 n + 5 mn 2 – 5 d) 16 x 4 – 1.
\(1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)
Do đó : \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot\left(n-1\right)}{2\cdot3\cdot4\cdot...\cdot n}\cdot\frac{3\cdot4\cdot5\cdot...\cdot\left(n+1\right)}{2\cdot3\cdot4\cdot...\cdot n}\)
\(=\frac{1}{n}\cdot\frac{n+1}{2}=\frac{n+1}{2n}\)