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ta có:\(192⋮2\Rightarrow192\times5⋮2\)
\(2018⋮2\)
\(\Rightarrow192\times5+2018⋮2\)
Ta có 192 . 5 là số chẵn mà 2018 là chẵn => 192 . 5 + 2018 chia hết cho 2
Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
Vậy ......................
~ Học tốt ~
Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Áp dụng BĐT Svác-xơ ta có:
\(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}\ge\frac{\left(\sqrt{2017}+\sqrt{2018}\right)^2}{\sqrt{2017}+\sqrt{2018}}=\sqrt{2017}+\sqrt{2018}\)
do \(\frac{2017}{\sqrt{2018}}\ne\frac{2018}{\sqrt{2017}}\)nên dấu "=" không xảy ra
Vậy \(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)
\(2018^{13}-2018^{12}=2018^{12}\left(2018-1\right)=2018^{12}.2017\)
\(2018^{11}.2018^{10}=2018^{12}.2018^9\)
Nhận thấy: \(2017< 2018^9\)=> \(2018^{12}.2017< 2018^{12}.2018^9\)
hay \(2018^{13}-2018^{12}< 2018^{11}.2018^{10}\)
Ta có :
\(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Ta thấy :
\(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\left(1\right)\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\left(2\right)\)
từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
trả lời
192.5+2018
=960+2018
=2978
chúc bạn học tốt!