Ta có \(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2018}{2^{2018}}+\frac{2019}{2^{2019}}\)

=> 2S = \(1+1+\frac{3}{2^2}+...+\frac{2018}{2^{2017}}+\frac{2019}{2^{2018}}\)

Khi đó 2S - S = \(\left(1+1+\frac{3}{2^2}+..+\frac{2018}{2^{2017}}+\frac{2019}{2^{2018}}\right)-\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2018}{2^{2018}}+\frac{2^{2019}}{2019}\right)\)

=> S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}-\frac{2019}{2^{2019}}\)

Đặt P = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}\)

=> 2P = \(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\)

Khi đó 2P - P = \(\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}\right)\)

P = \(2-\frac{1}{2^{2018}}\)

Thay P vào S 

=> S = \(2-\frac{1}{2^{2018}}-\frac{2019}{2^{2019}}=2-\frac{2}{2^{2019}}-\frac{2019}{2^{2019}}=2-\frac{2021}{2^{2019}}< 2\)

Vậy S < 2

quãng đường AB là : 40*5=200 km

thời gian khi ô tô chạy 500km/h là t= 200:500=0.4 h = 24 phút

21xy - 35x + 18y - 43 = 0

=> 7x(3y - 5) + 18y - 30 - 13 = 0

=> 7x(3y - 5) + 6(3y - 5) = 13

=> (7x + 6)(3y - 5) = 13

Vì \(x;y\inℤ\Rightarrow\hept{\begin{cases}7x+6\inℤ\\3y-5\inℤ\end{cases}}\)

Khi đó 13 = 1.13 = (-1).(-13)

Lập bảng xét các trường hợp 

Vậy x = 1 ; y = 2

ta có

\(21xy-35x+18x-30=13\)

\(\Leftrightarrow\left(3y-5\right)\left(7x+6\right)=13\)

do đó \(\hept{\begin{cases}3y-5=1\\7x+6=13\end{cases}}\)hoặc \(\hept{\begin{cases}3y-5=-1\\7x+6=-13\end{cases}}\)hoặc \(\hept{\begin{cases}3y-5=13\\7x+6=1\end{cases}}\)hoặc \(\hept{\begin{cases}3y-5=-13\\7x+6=-1\end{cases}}\)

tương ứng ta tìm \(\left(x,y\right)\in\left(1;2\right)\)là cặp giá trị nguyên duy nhất thỏa mãn.