\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)

\(=\frac{n}{2\left(3n+2\right)}\)

a) \(\lim \frac{{3n - 1}}{n} = \lim \frac{{n\left( {3 - \frac{1}{n}} \right)}}{n} = \lim \left( {3 - \frac{1}{n}} \right) = 3 - 0 = 3\)

b) \(\lim \frac{{\sqrt {{n^2} + 2} }}{n} = \lim \frac{{\sqrt {{n^2}\left( {1 + \frac{2}{{{n^2}}}} \right)} }}{n} = \lim \frac{{n\sqrt {1 + \frac{2}{{{n^2}}}} }}{n} = \lim \sqrt {1 + \frac{2}{{{n^2}}}}  = 1 + 0 = 1\)

c) \(\lim \frac{2}{{3n + 1}} = \lim \frac{2}{{n\left( {3 + \frac{1}{n}} \right)}} = \lim \left( {\frac{2}{n}.\frac{1}{{3 + \frac{1}{n}}}} \right) = \lim \frac{2}{n}.\lim \frac{1}{{3 + \frac{1}{n}}} = 0.\frac{1}{{3 + 0}} = 0\)

d) \(\lim \frac{{\left( {n + 1} \right)\left( {2n + 2} \right)}}{{{n^2}}} = \lim \frac{{n\left( {1 + \frac{1}{n}} \right).2n\left( {1 + \frac{1}{n}} \right)}}{{{n^2}}} = \lim \frac{{2{n^2}{{\left( {1 + \frac{1}{n}} \right)}^2}}}{{{n^2}}}\)

                                                      \( = \lim 2{\left( {1 + \frac{1}{n}} \right)^2} = 2.{\left( {1 + 0} \right)^2} = 2\)

\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right).\left(3n+2\right)}=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right).\left(3n+2\right)}\right)\)

                                                                          \(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)

                                                                            \(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)

                                                                              \(=\frac{1}{3}.\left(\frac{3n+2}{2.\left(3n+2\right)}-\frac{2}{2.\left(3n+4\right)}\right)\)

                                                                                \(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}=\frac{n}{2.\left(3n+2\right)}\)

Mk làm mẫu cho 1 phần rùi các câu còn lại làm tương tự nhé

a)    \(\frac{3n-2}{n-3}=3+\frac{7}{n-3}\)

Để   \(\frac{3n-2}{n-3}\)nguyên  thì   \(\frac{7}{n-3}\)nguyên

hay     \(n-3\)\(\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta lập bảng sau:

\(n-3\)     \(-7\)               \(-1\)                   \(1\)                    \(7\)

\(n\)              \(-4\)                  \(2\)                    \(4\)                   \(10\)

Vậy....

Em con quá non