Question

Answer 1

1: \(A=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)

2: =>\(\dfrac{5}{2}< \dfrac{6a+b}{6}< \dfrac{13}{3}\)

=>15<6a+b<26

=>6a+b\(\in\left\{16;17;18;19;20;21;22;23;24;25\right\}\)

=>6a=18 và \(b\in\left\{0;1;2;3;4;5;6;7\right\}\) hoặc \(6a=24;b\in\left\{0;1\right\}\)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=3\\0< =b< =7\end{matrix}\right.\\\left\{{}\begin{matrix}a=4\\0< =b< =1\end{matrix}\right.\end{matrix}\right.\)