Toán Lớp 6: e,K+1/2021+K+2/2020=K+3/2019 +K+4/2018 giupa mik vs

Question

Toán Lớp 6:  e,K+1/2021+K+2/2020=K+3/2019 +K+4/2018 giupa mik vs

ngochuong2k2 · Answer

Giải đáp: k=-2022 Lời giải và giải thích chi tiết: 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.

aivilanlan · Answer

$  $   (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