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A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
A=\(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
A=\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
A=\(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\)
B-A=\(\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
B-A=1/1008
Vì \(2015^{2016}+1< 2015^{2017}+1\Rightarrow\frac{2015^{2016}+1}{2015^{2017}+1}< 1\)
\(\Rightarrow A=\frac{2015^{2016}+1}{2015^{2017}+1}< \frac{2015^{2016}+1+2014}{2015^{2017}+1+2014}=\frac{2015\left(2015^{2015}+1\right)}{2015\left(2015^{2016}+1\right)}=\frac{2015^{2015}+1}{2015^{2016}+1}=B\)
Vậy \(A< B\)
\(2015A=\frac{2015^{2017}+2015}{2015^{2017}+1}=\frac{2015^{2017}+1+2014}{2015^{2017}+1}=1+\frac{2014}{2015^{2017}+1}\)
\(2015B=\frac{2015^{2016}+2015}{2015^{2016}+1}=\frac{2015^{2016}+1+2014}{2015^{2016}+1}=1+\frac{2014}{2015^{2016}+1}\)
vì \(\frac{2014}{2015^{2017}+1}< \frac{2014}{2015^{2016}+1}\)
nên \(2015A< 2015B\)
=> \(B>A\)
Ta có 20152015 = 20152015
Ta so sánh 20152016+1 và 20152011+1
Vì 20152016 > 20152011
=> 20152016+1 > 20152011 +1
2 phân số có cùng tử số, mẫu của phân số nào nhỏ hơn thì phân số đó lớn hơn
=>\(\frac{2015^{2015}+1}{2015^{2016}+1}
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
A = \(\frac{2015.2016-1}{2015.2016}\)= \(\frac{2015.2016}{2015.2016}\)\(-\)\(\frac{1}{2015.2016}\)= 1 \(-\)\(\frac{1}{2015.2016}\)
B = \(\frac{2016.2017-1}{2016.2017}\)= \(\frac{2016.2017}{2016.2017}\)\(-\)\(\frac{1}{2016.2017}\)= 1 \(-\)\(\frac{1}{2016.2017}\)
Vì \(\frac{1}{2015.2016}\)> \(\frac{1}{2016.2017}\)
=> 1 \(-\)\(\frac{1}{2015.2016}\)< \(1-\)\(\frac{1}{2016.2017}\)
=> A < B