Đặt: \(x^{673}=a;y^{673}=b\Rightarrow a^3=b^3-b^2-b+2\)

\(+,b=0\Rightarrow a^3=-2\left(\text{vô lí}\right)\)

\(+,b=1\Rightarrow a=1\left(\text{thỏa mãn}\right)\)

\(+,b=-1\Rightarrow a^3=3\left(\text{vô lí vì a nguyên}\right)\)

\(+,b=-2\Rightarrow a^3=8\Leftrightarrow a=2\left(\text{loại vì x;y không nguyên}\right)\)

\(+,b\ne1;0;-1;-2\Rightarrow\left(b-1\right)^3< b^3-b^2-b+2< b^3\left(\text{nên loại}\right)\)

bạn tự kết luận

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

Ta có: \(C=\dfrac{2019-2018}{2019+2018}\)

\(\Leftrightarrow C=\dfrac{\left(2019-2018\right)\left(2019+2018\right)}{\left(2019+2018\right)^2}\)

\(\Leftrightarrow C=\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}\)

Ta có: \(\left(2019+2018\right)^2=2019^2+2018^2+2\cdot2019\cdot2018\)

\(2019^2+2018^2=2019^2+2018^2+0\)

Do đó: \(\left(2019+2018\right)^2>2019^2+2018^2\)

\(\Leftrightarrow\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}< \dfrac{2019^2-2018^2}{2019^2+2018^2}\)

\(\Leftrightarrow C< D\)

Ta có: B = (2018 + 2019)/(2019 + 2020) = (2018 + 2019)/4039 = 2018/4039 + 2019/4039
Ta thấy : 2018/2019 > 2018/4039
            2019/2020 > 2019/4039
=> 2018/2019 + 2019/2020 > 2018/4039 > 2019/4039
=> 2018/2019 + 2019/2020 > (2018 + 2019)/(2019 + 2020)
=> A  > B

\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)

\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)

=>B<1

=>A>B