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1. \(G=2016.2016=\left(2014+2\right)\left(2018-2\right)=2014.2018-4028+4036-4=2014.2018+4\)
vì 2014.2018+4 >2014.2018
=> G>H
\(\frac{2016.2016}{2013.2019}=\frac{\left(2013+3\right)\left(2019-3\right)}{2013.2019}=\frac{2013.2019-6039+6057-9}{2013.2019}=\frac{2013.2019+9}{2013.2019}=1+\frac{9}{2013.2019}\)
vì \(1+\frac{9}{2013.2019}>1\)
\(\frac{2016.2016}{2013.2019}>1\)
S = 2017.2018.( 2016/2017 - 2016/2018 )
S= 2017.2018.2016/2017 - 2017.2018.2016/2018
S = 2018.2016 - 2016.2017
S = 2016.( 2018 - 2017 )
S = 2016.1 = 2016
a, Ta có: \(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)
\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)
\(\Rightarrow A=\dfrac{2016+2017}{2017+2018}< B=\dfrac{2016}{2017}+\dfrac{2017}{2018}\)
Vậy A < B
b, Ta có: \(\dfrac{2017}{2016+2017}< \dfrac{2017}{2016}\)
\(\dfrac{2018}{2016+2017}< \dfrac{2018}{2017}\)
\(\Rightarrow M=\dfrac{2017+2018}{2016+2017}< N=\dfrac{2017}{2016}+\dfrac{2018}{2017}\)
Vậy M < N
có B=2015+2016+\(\frac{2017}{2016}\)+2017+2018
B=\(\frac{2015}{2015+2016+2017}\)+\(\frac{2016}{2016+2017+2018}\)+\(\frac{2017}{2016+2017+2018}\)
vì \(\frac{2015}{2016}\)>\(\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}\)>\(\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}\)>\(\frac{2017}{2016+2017+2018}\)
⇒A>B
Chúc bạn học tốt :")
Dễ thấy B<1.
\(A=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)\)\(=3-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)\)
\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)
Vậy A>2.
Vậy A>B.
các bạn ơi chiều nay mik đi học rùi các bạn giải nhanh nha
\(\text{2016 × 2016 -2018 × 2014}\)
\(\text{= 2016 x ( 2014 + 2 ) - ( 2016 + 2 ) x 2014 }\)
\(\text{= ( 2016 x 2014 + 2016 x 2 ) - ( 2014 x 2016 + 2014 x 2 )}\)
\(\text{= 2016 x 2 - 2014 x 2 }\)
\(\text{= 4}\)
Đúng 100 % đó
k mk nha các bạn