\(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}\)
\(A=\dfrac{2015}{2016}-1+\dfrac{2016}{2017}-1+\dfrac{2017}{2018}-1+\dfrac{2018}{2015}-1+4\)
\(A=-\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{3}{2015}+4\)
\(A=\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)+\left(\dfrac{1}{2015}-\dfrac{1}{2017}\right)+\left(\dfrac{1}{2015}-\dfrac{1}{2018}\right)+4\)
Có: \(\dfrac{1}{2015}-\dfrac{1}{2016}>0\) ; \(\dfrac{1}{2015}-\dfrac{1}{2017}>0\) ; \(\dfrac{1}{2015}-\dfrac{1}{2018}>0\)
=> \(A=\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)+\left(\dfrac{1}{2015}-\dfrac{1}{2017}\right)+\left(\dfrac{1}{2015}-\dfrac{1}{2018}\right)+4>4\)
hay A>4
tham khảo
A=2015/2016+2016/2017+2017/2018+2018/2015
A=2015/2016+2016/2017+2017/2018+2018/2015
A=2016−1/2016+2017−1/2017+2018−1/2018+2015+3/2015
A=2016-12016+2017-12017+2018-12018+2015+32015
A=1−1/2016+1−1/2017+1−1/2018+1+3/2015
A=1-1/2016+1-1/2017+1-1/2018+1+3/2015
A=4+(1/2015−1/2016)+(1/2015−1/2017)+(1/2015−1/2018)
A=4+(1/2015-1/2016)+(1/2015-1/2017)+(1/2015-1/2018)
A=4+2016−2015/2015.2016+2017−2015/2015.2017+2018−2015/2015.2018
A=4+2016-20152015.2016+2017-20152015.2017+2018-20152015.2018
A=4+1/2015.2016+2/2015.2017+3/2015.2018
A=4+1/2015.2016+2/2015.2017+3/2015.2018
A>4+0+0+0=4
A = 2015/2016 + 2016/2017 + 2017/2018 + 2018/2015
A = 2015/2016 + 2016/2017 + 2017/2018 + 2018/2015
A = 2016 − 1/2016 + 2017 − 1/2017 + 2018 − 1/2018 + 2015 + 3/2015
A = 2016 - 12016 + 2017 - 12017 + 2018 - 12018 + 2015 + 32015
A = 1 − 1/2016 + 1 − 1/2017 + 1 − 1/2018 + 1 + 3/2015
A = 1 - 1/2016 + 1 - 1/2017 + 1 - 1/2018 + 1 + 3/2015
A = 4 + (1/2015 − 1/2016) + (1/2015 − 1/2017) + (1/2015 − 1/2018)
A = 4+(1/2015-1/2016)+(1/2015-1/2017)+(1/2015-1/2018)
A = 4 + 2016 − 2015/2015 . 2016 +2017 − 2015/2015 . 2017 + 2018 − 2015/2015 . 2018
A = 4 + 2016 - 2015 . 2016 + 2017 - 2015 . 2017 + 2018 - 2015 . 2018
A = 4 + 1/2015 . 2016 + 2/2015 . 2017 + 3/2015 . 2018
A > 4 + 0 + 0 + 0 = 4
\(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}\)
\(A=\dfrac{2016-1}{2016}+\dfrac{2017-1}{2017}+\dfrac{2018-1}{2018}+\dfrac{2015+1+1+1}{2015}\)
\(A=1-\dfrac{1}{2016}+1-\dfrac{1}{2017}+1-\dfrac{1}{2018}+1+\dfrac{1}{2015}+\dfrac{1}{2015}+\dfrac{1}{2015}\)
\(A=4+\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)+\left(\dfrac{1}{2015}-\dfrac{1}{2017}\right)+\left(\dfrac{1}{2015}-\dfrac{1}{2018}\right)\)
Vì: \(\dfrac{1}{2015}>\dfrac{1}{2016}>\dfrac{1}{2017}>\dfrac{1}{2018}\)
\(\Rightarrow\dfrac{1}{2015}-\dfrac{1}{2016}>0;\dfrac{1}{2015}-\dfrac{1}{2017}>0;\dfrac{1}{2015}-\dfrac{1}{2018}>0\)
\(\Rightarrow A>4\) (đpcm)
Ai bik giải bài này giúp mik với
