e,K+1/2021+K+2/2020=K+3/2019 +K+4/2018 giupa mik vs

nhat9983

2 năm trước

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duongngohanh2405

$\\$

`(k+1)/2021 + (k+2)/2020 = (k+3)/2019 + (k+4)/2018`

`-> (k+1)/2021 + (k+2)/2020 + 2 = (k+3)/2019 + (k+4)/2018 + 2`

`-> ( (k+1)/2021 +1) + ( (k+2)/2020 + 1) = ( (k+3)/2019 + 1) + ( (k+4)/2018 + 1)`

`-> ( (k+1)/2021 + 2021/2021) + ( (k+2)/2020 + 2020/2020) = ( (k+3)/2019 + 2019/2019) + ( (k+4)/2018 + 2018/2018)`

`-> (k+1+2021)/2021 + (k+2+2020)/2020 = (k+3+2019)/2019 + (k+4+2018)/2018`

`-> (k+2022)/2021 + (k+2022)/2020 = (k+2022)/2019 + (k+2022)/2018`

`-> (k+2022)/2021 + (k+2022)/2020 - (k+2022)/2019 - (k+2022)/2018 =0`

`-> (k+2022) (1/2021 + 1/2020 - 1/2019 - 1/2018)=0`

`-> k+2022=0` (Do `1/2021 + 1/2020 - 1/2019 - 1/2018 \ne 0`)

`->k=0-2022`

`-> k=-2022`

Vậy `k=-2022`

Vote (0) Phản hồi (0) 2 năm trước

146752240508053

Đáp án:

`k=-2022`

Giải thích các bước giải:

`e)(k+1)/2021+(k+2)/2020=(k+3)/2019+(k+4)/2018`

`=>(k+1)/2021+1+(k+2)/2020+1=(k+3)/2019+1+(k+4)/2018+1`

`=>(k+2022)/2021+(k+2022)/2020=(k+2022)/2019+(k+2022)/2018`

`=>(k+2022)(1/2021+1/2020-1/2019-1/2018)=0`

Vì `1/2021<1/2019`

`\qquad 1/2020<1/2018`

`=>1/2021+1/2020-1/2019-1/2018<0`

`=>k+2022=0`

`=>k=-2022`

Vậy với `k=-2022` thì `(k+1)/2021+(k+2)/2020=(k+3)/2019+(k+4)/2018`.

Vote (0) Phản hồi (0) 2 năm trước

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