\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)

\(P=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)

\(P=1+\frac{1}{2}.3+\frac{1}{3}.6+\frac{1}{4}.10+....+\frac{1}{2016}.2033136\)

\(P=1+\frac{3}{2}+4+\frac{5}{2}+....+\frac{2017}{2}\)

\(P=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2017}{2}\)

\(P=\frac{2+3+4+5+....+2017}{2}=\frac{2035152}{2}=1017576\)

\(\left(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\right):\left(1+2+3+...+2016\right)\\ =\left(1-1+1-1\right):\left(1+2+3+...+2016\right)\\ =0:\left(1+2+3+...+2016\right)=0\)

Mk ko bt t mình nhé mk mới giam gia thôi

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2015.2016.2017}\)

\(A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2017-2015}{2015.2016.2017}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)

\(2A=\frac{1}{1.2}-\frac{1}{2016.2017}\)

\(A=\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\div2\)

khó quá mình ko biết làm

khjk63201 \'oiuo

mìh k biết đâu tại mình an nhằm thứ lỗi nhé

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

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

\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)

\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)

Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)

\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{2016}\left(1+2+...+2016\right)\)\(=1+\dfrac{2.3}{2.2}+\dfrac{3.4}{3.2}+...+\dfrac{2016.2017}{2016.2}\)

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

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

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

Đặt \(A=2+3+...+2017\)

\(=2017+2016+...+2\)

\(\Rightarrow2A=\left(2+2017\right)+\left(3+2016\right)+...+\left(2017+2\right)\) ( 2016 cặp số )

\(\Rightarrow2A=2019+2019+...+2019\) ( 2016 số )

\(\Rightarrow2A=4070304\)

\(\Rightarrow A=2035152\)

\(\Rightarrow P=1017576\)

Vậy...

P= 1+1/2.3+1/3.6+...+1/2016.2033136

P= 1+3/2+2+...+2017/2

P= 2/2+3/2+4/2+...+2017/2

P=\(\dfrac{2+3+4+...+2017}{2}\)

P= \(\dfrac{2035152}{2}\)

P= 1017576

Anh Nguyễn Huy Tú làm đúng như hơi gọn;mình sẽ làm lại đầy đủ hơn nhé.

\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+....+\dfrac{1}{2016}\left(1+....+2016\right)\\ =1+\dfrac{1}{2}.\dfrac{\left(1+2\right).2}{2}+\dfrac{1}{3}.\dfrac{\left(1+3\right).3}{2}+...+\dfrac{1}{2016}.\dfrac{\left(2016+1\right).2016}{2}\\ =\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+....+\dfrac{2017}{2}=\dfrac{\left(2+3+....+2017\right)}{2}=\dfrac{\left(2017+2\right).2016}{2}=2009.1008=2025072\)