$\frac{x+19}{2019}$ $\frac{x+18}{2018}$ $\frac{x+17}{2017}$ +…..+ $\frac{x+1}{2001}$ =19 01/09/2021 Bởi Samantha $\frac{x+19}{2019}$ $\frac{x+18}{2018}$ $\frac{x+17}{2017}$ +…..+ $\frac{x+1}{2001}$ =19
Đáp án + Giải thích các bước giải: `(x+19)/2019 + (x+18)/2018 + (x+17)/2017 + … + (x+1)/2001 = 19` `to (x+19)/2019 + (x+18)/2018 + (x+17)/2017 + … + (x+1)/2001 – 19 = 0` `to ( (x+19)/2019 – 1 ) + ( (x+18)/2018 – 1 ) + ( (x+17)/2017 – 1 ) + … + ( (x+1)/2001 – 1 ) = 0 ` `to (x+19-2019)/2019 + (x+18-2018)/2018 + (x+17-2017)/2017 + … + (x+1-2001)/2001 = 0` `to (x-2000)/2019 + (x-2000)/2018 + (x-2000)/2017 + … + (x-2000)/2001 = 0` `to (x-2000) . (1/2019+1/2018+1/2017+…+1/2001)=0` Mà `1/2019+1/2018+1/2017+…+1/2001 ne 0` `to x-2000=0` `to x=2000` Vậy `x=2000` Bình luận
(x+19/2019 – 1)+(x+18/2018 – 1)+(x+2017/2017 – 1)+…+(x+1/2001 – 1)=0 (Đổi 19 sang vế trái) => x-2000/2019 + x-2000/2018 + x-2000/2017+…+ x-2000/2001 =0 nhóm nhân tử chung x-2000: => (x-2000)*(1/2019 +1/2018 +1/2017+…+ 1/2001)=0 vì 1/2019+1/2018+1/2017+…+1/2001 >0 =>x-2000=0 =>x=2000 Bình luận
Đáp án + Giải thích các bước giải:
`(x+19)/2019 + (x+18)/2018 + (x+17)/2017 + … + (x+1)/2001 = 19`
`to (x+19)/2019 + (x+18)/2018 + (x+17)/2017 + … + (x+1)/2001 – 19 = 0`
`to ( (x+19)/2019 – 1 ) + ( (x+18)/2018 – 1 ) + ( (x+17)/2017 – 1 ) + … + ( (x+1)/2001 – 1 ) = 0 `
`to (x+19-2019)/2019 + (x+18-2018)/2018 + (x+17-2017)/2017 + … + (x+1-2001)/2001 = 0`
`to (x-2000)/2019 + (x-2000)/2018 + (x-2000)/2017 + … + (x-2000)/2001 = 0`
`to (x-2000) . (1/2019+1/2018+1/2017+…+1/2001)=0`
Mà `1/2019+1/2018+1/2017+…+1/2001 ne 0`
`to x-2000=0`
`to x=2000`
Vậy `x=2000`
(x+19/2019 – 1)+(x+18/2018 – 1)+(x+2017/2017 – 1)+…+(x+1/2001 – 1)=0 (Đổi 19 sang vế trái)
=> x-2000/2019 + x-2000/2018 + x-2000/2017+…+ x-2000/2001 =0
nhóm nhân tử chung x-2000:
=> (x-2000)*(1/2019 +1/2018 +1/2017+…+ 1/2001)=0
vì 1/2019+1/2018+1/2017+…+1/2001 >0
=>x-2000=0
=>x=2000