\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\left(\dfrac{2015}{2}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)+1}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}=\dfrac{1}{2017}\)

Xét phần mẫu số: \(\frac{2016}{1}\) = 2016 = 1 + 1 + 1 +...+ 1 (2016 số hạng 1)

Ta có: (1+\(\frac{2015}{2}\)) + (1+\(\frac{2014}{3}\)) + (1+\(\frac{2013}{4}\)) + ... + (1+\(\frac{1}{2016}\))

= \(\frac{2017}{2}\) + \(\frac{2017}{3}\) + \(\frac{2017}{4}\) + ... + \(\frac{2017}{2016}\)

= 2016 ^x (\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+...+\(\frac{1}{2016}\))

=> \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}{2016x\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}\) 

Rút \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\) ở cả tử số và mẫu số, ta còn lại \(\frac{1}{2016}\)

Vậy \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}}\) = \(\frac{1}{2016}\)

sao mà khó thế !!!!!!!!!!!!

Đề bài có bị sai ko mà ở phần tử số là \(\frac{1}{2006}\) mà mẫu số lại là \(\frac{1}{2016}\)?

Còn cần không:v

bằng 15 hay sao ý

\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\) 

                                                 =\(\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}\)

                                                    =\(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

áp dụng vào biểu thức ta có\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

                       =\(1-\frac{1}{\sqrt{2016}}\)

   đến đây cậu tự giải nốt nhé

bạn coi thử sách VHB đi hình như có đấy

mình ko có sách đấy 

\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

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

\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=2018\)

Ta có : 

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

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

\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

\(A=2018\)

Vậy \(A=2018\)

Chúc bạn học tốt ~ 

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