Q10 là 10.Q????
1) Ta có: \(\frac{n+3}{2n-2}\)\(=\frac{n+3}{2\left(n-1\right)}\)
Để \(\frac{n+3}{2\left(n-1\right)}\in Z\) thì \(\left\{{}\begin{matrix}n+3\in Z\Rightarrow n\in Z\\2\left(n-1\right)\in Z\Rightarrow n\in Z\\2\left(n-1\right)\ne0\Rightarrow n-1\ne0\Rightarrow n\ne1\end{matrix}\right.\)
Vậy để \(\frac{n+3}{2n-2}\in Z\) thì: \(n\in Z;n\ne1\)
mà n là số tự nhiên \(\Rightarrow n\in N;n\ne1\)
Vậy: Để \(\frac{n+3}{2n-2}\in Z\) thì: n\(\in N;n\ne1\)
2) Ta có: \(Q=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(\Leftrightarrow Q=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Leftrightarrow Q=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)
\(\Leftrightarrow10.Q=10.\frac{1}{10}=1\)