Giải:

Ta có:

\(P=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)

và \(Q=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)

Vì \(\left\{{}\begin{matrix}\dfrac{2016}{2017}=\dfrac{2016}{2017}\\\dfrac{2017}{2018}=\dfrac{2017}{2018}\\\dfrac{2018}{2019}=\dfrac{2018}{2019}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)

Hay \(P=Q\)

Vậy ...

\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)

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

A=1-\(\frac{1}{2017}\)+1-\(\frac{1}{2018}\)+1-\(\frac{1}{2019}\)+1+\(\frac{3}{2016}\)

Dạng bài tương tự như bài này, bạn áp dụng cách làm vào làm bài của bạn nhé: Câu hỏi của Dao Dao - Toán lớp 7 | Học trực tuyến

\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2018}}{\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}}\)

Đặt \(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\) là B

\(B=\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\\ =\dfrac{2017}{1}+1+\dfrac{2016}{2}+1+...+\dfrac{1}{2017}+1-2017\\ =\dfrac{2018}{1}+\dfrac{2018}{2}+...+\dfrac{2018}{2017}-2017\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\left(2018-2017\right)\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+1\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\\ =2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)

\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2018}}{2018\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}{2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017}{2018}\)

Mơn cac ban, minh dang can cau nay ^^

Đề ra là thế này hả: So sánh A và B ....................

Vì hai phân số của A đều bé hơn 1 nên tổng chúng bé hơn 2. Vậy A < B

k mk đi mà làm ơnnnnnnnnnn

Tính A và B rồi ta đi so sánh:

A = \(\frac{2016}{2017}\) + \(\frac{2017}{2018}\) = \(1.999008674\)

B = \(\frac{2016+2017}{2017+2018}\) = \(0.9995043371\)

Mà 1.999008674 > 0.9995043371

Nên: A > B

Giải như bạn Trần Nhật Quỳnh thà không làm còn hơn.

Ta có :

\(M=\dfrac{2018^{2017}+1}{2018^{2018}+1}< 1\)

\(\Rightarrow M< \dfrac{2018^{2017}+1+2017}{2017^{2018}+1+2017}=\dfrac{2018^{2017}+2018}{2017^{2018}+2018}=\dfrac{2018\left(2018^{2016}+1\right)}{2018\left(2018^{2017}+1\right)}=\dfrac{2018^{2016}+1}{2018^{2017}+1}=N\)

\(\Rightarrow M< N\)

Giải:

Ta có:

\(2018M=\dfrac{\left(2018^{2017}+1\right)2018}{2018^{2018}+1}.\)

\(2018M=\dfrac{2018^{2018}+2018}{2018^{2018}+1}.\)

\(2018M=\dfrac{\left(2018^{2018}+1\right)+2017}{2018^{2018}+1}.\)

\(2018M=\dfrac{2018^{2018}+1}{2018^{2018}+1}+\dfrac{2017}{2018^{2018}+1}.\)

\(2018M=1+\dfrac{2017}{2018^{2018}+1}._{\left(1\right)}\)

Ta lại có:

\(2018N=\dfrac{\left(2018^{2016}+1\right)2018}{2018^{2017}+1}.\)

\(2018N=\dfrac{2018^{2017}+2018}{2018^{2017}+1}.\)

\(2018N=\dfrac{\left(2018^{2017}+1\right)+2017}{2018^{2017}+1}.\)

\(2018N=\dfrac{2018^{2017}+1}{2018^{2017}+1}+\dfrac{2017}{2018^{2017}+1}.\)

\(2018N=1+\dfrac{2017}{2018^{2017}+1}._{\left(2\right)}\)

Và \(\dfrac{2017}{2018^{2018}+1}< \dfrac{2017}{2018^{2017}+1}._{\left(3\right)}\)

Từ \(_{\left(1\right);\left(2\right)}\) và \(_{\left(3\right)}\Rightarrow2018M< 2018N\Rightarrow M< N.\)

Vậy......

~ Học tốt!!! ~

tớ chịu thôi SORRY CẬU RẤT NHIỀU

A=2015/2016+2016/2017+2017/2018>2015/2018+2016/2018+2017/2018

=6048/2018>1

B=2015+2016+2017/2016+2017+2018=6048/6051<1

=>A>B