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Đặt 2018=a

\(VT=a\left(a-1\right)\left(a+1\right)=a\left(a^2-1\right)=a^3-a< a^3\)

Do đó: VT<VP

1 tháng 9 2020

a) Ta có : \(\frac{-3}{100}< 0< \frac{2}{3}\)

\(\Rightarrow\frac{-3}{100}< \frac{2}{3}\)

b) Ta có : \(\frac{267}{268}< 1< \frac{1347}{1343}\)

\(\Rightarrow\frac{267}{268}< \frac{1347}{1343}\)

\(\Rightarrow\frac{267}{-268}< \frac{-1347}{1343}\)

c) Ta có : \(\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)

                 \(\frac{2018.2019-1}{2018.2019}=\frac{2018.2019}{2018.2019}-\frac{1}{2018.2019}=1-\frac{1}{2018.2019}\)

mà \(2017.2018< 2018.2019\)

\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)

\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

\(\Rightarrow\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)

d) Ta có : \(\frac{2017.2018}{2017.2018+1}=\frac{2017.2018+1}{2017.2018+1}-\frac{1}{2017.2018+1}=1-\frac{1}{2017.2018+1}\)

                 \(\frac{2018.2019}{2018.2019+1}=\frac{2018.2019+1}{2018.2019+1}-\frac{1}{2018.2019+1}=1-\frac{1}{2018.2019+1}\)

mà \(2017.2018+1< 2018.2019+1\)

\(\Rightarrow\frac{1}{2017.2018+1}>\frac{1}{2018.2019+1}\)

\(\Rightarrow1-\frac{1}{2017.2018+1}< 1-\frac{1}{2018.2019+1}\)

\(\Rightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}\)

 

14 tháng 10 2019

2017.2019 = (2018-1)(2018+1) = 20182 -1 => a =1

b= 20183 +1 (???)

15 tháng 7 2018

\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}=\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)+\left(1-\frac{1}{2019}\right)\)

\(A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)< 3\)

15 tháng 7 2018

Ta có :

2016/2017 < 1

2017/2018 < 1

2018/2019 < 1

Mà 2016/2017 + 2017/2018 + 2018/2019 < 1 + 1 + 1 = 3

Nên A < 3

15 tháng 7 2018

\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)

Ta có:

 \(\frac{2016}{2017}< 1\)

\(\frac{2017}{2018}< 1\)

\(\frac{2018}{2019}< 1\)

\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}< 1+1+1=3\)

\(\Rightarrow A< 3\)

Vậy \(A< 3\)

Tham khảo nhé

15 tháng 7 2018

\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)

\(=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}\)

\(=\left(1+1+1\right)-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

\(=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)< 3\)

Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}< 3\left(đpcm\right)\)

11 tháng 7 2018

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  \(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)

7 tháng 8 2017

Bằng nhau nha

link nà:https://olm.vn/hoi-dap/tim-kiem?q=so+s%C3%A1nh+:+A=2017%5E2017/2018%5E2017+1B=2017%5E2016+1/2017%5E2017+1+&id=862033

21 tháng 4 2018

Thanks