\(S=\dfrac{1}{2018}\left(1+\dfrac{1}{1}+1+\dfrac{1}{2}+1+\dfrac{1}{3}+...+1+\dfrac{1}{2018}\right)\)

\(S=\dfrac{1}{2018}\left(2018+\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)

\(S=1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)

Do \(\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2018}\right)>0\Rightarrow S>1\) (1)

Lại có:

\(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}< \dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+...+\dfrac{1}{1}=2018\)

\(\Rightarrow1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)< 1+\dfrac{1}{2018}.2018=2\)

\(\Rightarrow S< 2\) (2)

Từ (1), (2) \(\Rightarrow1< S< 2\)

\(\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải là số tự nhiên

S=1/2018

a)

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}\)

P/s: Cj chỉ biết làm ý a thôi nhé! Có j ko hiểu cmt nhé!

a: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)

b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)

\(\dfrac{2}{3}A=\dfrac{2}{3}-\left(\dfrac{2}{3}\right)^2+\left(\dfrac{2}{3}\right)^3-...+\left(\dfrac{2}{3}\right)^{2019}-\left(\dfrac{2}{3}\right)^{2020}\)

=>\(\dfrac{5}{3}A=1-\left(\dfrac{2}{3}\right)^{2020}=1-\dfrac{2^{2020}}{3^{2020}}=\dfrac{3^{2020}-2^{2020}}{3^{2020}}\)

=>\(A=\dfrac{3^{2020}-2^{2020}}{3^{2020}}:\dfrac{5}{3}=\dfrac{3^{2020}-2^{2020}}{5\cdot3^{2019}}\) ko là số nguyên

::((

so hard

\(A=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2019}+\left(\dfrac{1}{3}\right)^{2020}\)

\(\Rightarrow\dfrac{1}{3}A=\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2021}\)

\(\Rightarrow\dfrac{2}{3}A=A-\dfrac{1}{3}A=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{2020}-\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^{2021}=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^{2021}< \dfrac{1}{3}\)

\(\Rightarrow A< \dfrac{1}{2}\)

1)

DKCĐ: a>0,\(a\ne1\)

\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)