Toán chứng minh rằng 1/101+1/102+….+1/199+1/200>7/12 07/10/2021 By Lyla chứng minh rằng 1/101+1/102+….+1/199+1/200>7/12
Đáp án: Ta có: Gọi A=1/101+1/102+….+1/199+1/200 A=(1/101+1/102+…..+1/150)+(1/151+1/152+…+1/200) mà 1/101>1/150;1/102>150;…..;1/149>150 1/151>1/200;1/152>1/200;…..;1/199>1/200 =>A>(1/150 x 50)+(1/200 x 50) A>1/3+1/4=4/12+3/12=7/12 Vậy A>7/12 XIN HAY NHẤT NHA Trả lời
Ta có `1/101+1/102+…+1/199+1/200` `=(1/101+1/102+…+1/150)+(1/151+1/152+…+1/200)` Ta thấy `1/101>1/150;1/102>1/150;…;1/149>1/150` `=>1/101+1/102+…+1/150>1/150+1/150+…+1/150` (Có `50` số hạng) `=>1/101+1/102+…+1/150>1/3` `(1)` Ta thấy `1/151>1/200;1/152>1/200;…;1/199>1/200` `=>1/151+1/152+…+1/200>1/200+1/200+…+1/200` (Có `50` số hạng) `=>1/151+1/152+…+1/200>1/4` `(2)` Từ `(1)` và `(2)` ta có: `1/101+1/102+…+1/199+1/200>1/3+1/4=7/12` Vậy `1/101+1/102+…+1/199+1/200>7/12`. Trả lời
Đáp án:
Ta có:
Gọi A=1/101+1/102+….+1/199+1/200
A=(1/101+1/102+…..+1/150)+(1/151+1/152+…+1/200)
mà 1/101>1/150;1/102>150;…..;1/149>150
1/151>1/200;1/152>1/200;…..;1/199>1/200
=>A>(1/150 x 50)+(1/200 x 50)
A>1/3+1/4=4/12+3/12=7/12
Vậy A>7/12
XIN HAY NHẤT NHA
Ta có `1/101+1/102+…+1/199+1/200`
`=(1/101+1/102+…+1/150)+(1/151+1/152+…+1/200)`
Ta thấy `1/101>1/150;1/102>1/150;…;1/149>1/150`
`=>1/101+1/102+…+1/150>1/150+1/150+…+1/150` (Có `50` số hạng)
`=>1/101+1/102+…+1/150>1/3` `(1)`
Ta thấy `1/151>1/200;1/152>1/200;…;1/199>1/200`
`=>1/151+1/152+…+1/200>1/200+1/200+…+1/200` (Có `50` số hạng)
`=>1/151+1/152+…+1/200>1/4` `(2)`
Từ `(1)` và `(2)` ta có:
`1/101+1/102+…+1/199+1/200>1/3+1/4=7/12`
Vậy `1/101+1/102+…+1/199+1/200>7/12`.