\(B=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left(1-\dfrac{1}{10}\right)\cdot\left(1-\dfrac{1}{15}\right)\cdot...\cdot\left(1-\dfrac{1}{780}\right)\)

\(B=\left(1-\dfrac{1}{3+6+10+15+...+780}\right)\)

\(B=\left(1-\dfrac{1}{\left(780-3\right)\div3+1}\right)\)

\(B=\left(1-\dfrac{1}{260}\right)\)

\(B=\dfrac{259}{260}\)

\(S=\dfrac{1}{7^2}+\dfrac{2}{7^3}+\dfrac{3}{7^3}+...+\dfrac{69}{7^{70}}\)

\(S=\dfrac{1+2+3+...+69}{\left(7\right)^{2+3+4+...+70}}\)

\(S=\dfrac{\left(69-1\right)+1}{\left(7\right)^{\left(70-2\right)+1}}\)

\(S=\dfrac{69}{7^{69}}\)

\(\Rightarrow S=7\)

Vậy \(S< \dfrac{1}{36}\)

\(A=\dfrac{2}{1}+2+\dfrac{5}{1}+2+3+\dfrac{9}{1}+2+3+4+...+\dfrac{2041210}{1}+2+3+4+...+2020\)

\(A=\left(\dfrac{2+5+9+...+2041210}{1}\right)+2+3+4+...+2020\)

\(A=\left(\dfrac{\left(2041210-2\right)\div4+1}{1}\right)+2+3+4+...+2020\)

\(A=\dfrac{510304}{1}+\left(2+3+4+...+2020\right)\)

\(A=510304+\left(2020-2\right)+1\)

\(A=510304+2019\)

\(A=512323\)

\(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2020}\)

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

\(=\dfrac{1}{\left(2020-1\right)+1}\)

\(=\dfrac{1}{2020}\)

\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2022^2}\)

\(C=\dfrac{1}{\left(2+3+4+...+2022\right)^2}\)

\(C=\dfrac{1}{\left(2022-2+1\right)^2}\)

\(C=\dfrac{1}{2021^2}\)

\(C=\dfrac{1}{2021\cdot2021}\)

\(C=\dfrac{1\div2021}{2021}\)

\(C=1\)

Vì \(1>\dfrac{13}{18}\)

\(\Rightarrow C>\dfrac{13}{18}\)

có nha em

lx rồi nha em

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< 1\)

\(A=\dfrac{1}{\left(2+3+4+...+2021\right)^2}< 1\)

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

\(A=\dfrac{1}{\left(2020\right)^2}< 1\)

\(A=\dfrac{1}{2020\cdot2020}< 1\)

\(A=\dfrac{1}{2020}< 1\)

\(\left(1-\dfrac{1}{2^2}\right)\cdot\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{10^2}\right)\)

\(=\left(1-\dfrac{1}{\left(2\cdot3\cdot...\cdot10\right)^2}\right)\)

\(=\left(1-\dfrac{1}{\left(10-2+1\right)^2}\right)\)

\(=\left(1-\dfrac{1}{9^2}\right)\)

\(=1-\dfrac{1}{18}\)

\(=\dfrac{18}{18}+\dfrac{1}{18}\)

\(=\dfrac{17}{18}\)

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