-8/9+1/9.2/9+1/9.7/9 A=1/ 2 ²+1/3 ²+1/4 ²…1/9 ² chứng tỏ 8/9>A>2/5 03/09/2021 Bởi Valentina -8/9+1/9.2/9+1/9.7/9 A=1/ 2 ²+1/3 ²+1/4 ²…1/9 ² chứng tỏ 8/9>A>2/5
`-8/9+ 1/9 . 2/9 + 1/9 . 7/9` `= -8/9 + 1/9( 2/9+ 7/9)` `= -8/9 + 1/9 .1` `= -8/9 + 1/9` `= -7/9` Ta có: `1/2^2 < 1/1.2 ; 1/3^2 < 1/2.3; ….. ; 1/9^2 < 1/8.9` `=> 1/2^2 + 1/3^2 + 1/4^2+………+ 1/9^2 < 1/1.2 + 1/2.3 + …. + 1/8.9` `=> A < 1/1 -1/2 + 1/2 -1/3 +….+ 1/8- 1/9` `=>A < 1/1 -1/9` `=> A < 8/9 (1)` Lại có: `1/2^2 > 1/2.3 + 1/3^2 > 1/3.4 ; ………; 1/9^2 > 1/9.10` `=> 1/2^2 + 1/3^2 + 1/4^2+………+ 1/9^2 > 1/2.3 + 1/3.4 +…+ 1/9.10` `=> A > 1/2 -1/3 + 1/3 -1/4 +….+ 1/9-1/10` `=> A > 1/2 -1/10 = 2/5` `(2)` Từ `(1)` và` (2) “=> 8/9 > A > 2/5` Vậy` 8/9 > A > 2/5` Bình luận
$A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{9^2}\\ A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\cdots+\dfrac{1}{9.10}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{2}-\dfrac{1}{10}\\ =\dfrac{2}{5}\\ A<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\cdots+\dfrac{1}{8.9}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{8}-\dfrac{1}{9}\\ =\dfrac{1}{1}-\dfrac{1}{9}\\ =\dfrac{8}{9}$ Vậy $\dfrac{8}{9}>A>\dfrac{2}{5}$ Bình luận
`-8/9+ 1/9 . 2/9 + 1/9 . 7/9`
`= -8/9 + 1/9( 2/9+ 7/9)`
`= -8/9 + 1/9 .1`
`= -8/9 + 1/9`
`= -7/9`
Ta có: `1/2^2 < 1/1.2 ; 1/3^2 < 1/2.3; ….. ; 1/9^2 < 1/8.9`
`=> 1/2^2 + 1/3^2 + 1/4^2+………+ 1/9^2 < 1/1.2 + 1/2.3 + …. + 1/8.9`
`=> A < 1/1 -1/2 + 1/2 -1/3 +….+ 1/8- 1/9`
`=>A < 1/1 -1/9`
`=> A < 8/9 (1)`
Lại có: `1/2^2 > 1/2.3 + 1/3^2 > 1/3.4 ; ………; 1/9^2 > 1/9.10`
`=> 1/2^2 + 1/3^2 + 1/4^2+………+ 1/9^2 > 1/2.3 + 1/3.4 +…+ 1/9.10`
`=> A > 1/2 -1/3 + 1/3 -1/4 +….+ 1/9-1/10`
`=> A > 1/2 -1/10 = 2/5` `(2)`
Từ `(1)` và` (2) “=> 8/9 > A > 2/5`
Vậy` 8/9 > A > 2/5`
$A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{9^2}\\ A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\cdots+\dfrac{1}{9.10}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{2}-\dfrac{1}{10}\\ =\dfrac{2}{5}\\ A<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\cdots+\dfrac{1}{8.9}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots+\dfrac{1}{8}-\dfrac{1}{9}\\ =\dfrac{1}{1}-\dfrac{1}{9}\\ =\dfrac{8}{9}$
Vậy $\dfrac{8}{9}>A>\dfrac{2}{5}$