So sánh : `M = {5^2018 + 1} / {5 ^ 2017 + 1}` ` N = {5^ 2017 + 1} / { 5^2016 +1 } `

0 bình luận về “So sánh : `M = {5^2018 + 1} / {5 ^ 2017 + 1}` ` N = {5^ 2017 + 1} / { 5^2016 +1 } `”

1. Ta có : `M = ( 5^2018 + 1 )/( 5^2017 + 1 )`

   `1/5 . M = ( 5^2018 + 1 )/( 5 . ( 5^2017 + 1 ))`

   `1/5 . M = ( 5^2018 + 5 – 4 )/( 5^2018 + 5 )`

   `1/5 . M = 1 – 4/( 5^2018 + 5 )`

   Lại có : `N = ( 5^2017 + 1 )/( 5^2016 + 1 )`

   `1/5 . N = ( 5^2017 + 1 )/( 5 . ( 5^2016 + 1 ))`

   `1/5 . N = ( 5^2017 + 5 – 4 )/( 5^2017 + 5 )`

   `1/5 . N = 1 – 4/( 5^2017 + 5 )`

   Mà `4/( 5^2018 + 5 ) < 4/( 5^2017 + 5 ) ⇒ 1/5 . M > 1/5 . N ⇒ M > N`

   Vậy , `M > N`

    

   Bình luận

2. ` M = (5^(2018)+1)/(5^(2017)+1)`

   `\to 1/5 M = ( 5^(2018) +1)/ (5^(2018) +5) = ( 5^(2018) + 5 – 4)/(5^(2018)+5) = 1 – 4/(5^(2018)+5)`

   ` N = (5^(2017)+1)/(5^(2016)+1)`

   ` \to 1/5 N = ( 5^(2017) +1)/ (5^(2017) +5) = ( 5^(2017) + 5 – 4)/(5^(2017)+5) = 1 – 4/(5^(2017)+5)`

   Ta có ` 5^(2018)+5 > 5^(2017) +5`

   `\to 4/(5^(2018)+5) < 4/(5^(2017)+5) \to 1 – 4/(5^(2018)+5) > 1 – 4/(5^(2017)+5)`

   `\to 1/5 M > 1/5 N`

   `\to M > N`

    

   Bình luận