a) Ta xét các trường hợp:

+)  Với n = 3k  \(\left(k\in Z\right)\), ta có \(\left(n-1\right)\left(n+2\right)+12=\left(3k-1\right)\left(3k+2\right)+12\)

Ta thấy (3k - 1)(3k + 2) không chia hết cho 3, 12 chia hết cho 3 nên (3k - 1)(3k + 2) + 12 không chia hết cho 3 hay (3k - 1)(3k + 2) + 12 không chia hết cho 9.

+)  Với n = 3k + 1 \(\left(k\in Z\right)\), ta có \(\left(n-1\right)\left(n+2\right)+12=3k\left(3k+3\right)+12=9k\left(k+1\right)+12\)

Ta thấy \(9k\left(k+1\right)⋮9;12⋮̸9\Rightarrow9k\left(k+1\right)+12⋮̸9\)

+) Với n = 3k + 2 \(\left(k\in Z\right)\), ta có: \(\left(n-1\right)\left(n+2\right)+12=\left(3k+1\right)\left(3k+4\right)+12\)

Ta thấy (3k + 1)(3k + 4) không chia hết cho 3, 12 chia hết cho 3 nên (3k + 1)(3k + 4) + 12 không chia hết cho 3 hay (3k + 1)(3k + 4) + 12 không chia hết cho 9.

b) Tương tự bài trên.

Đặt :

\(A=\dfrac{3}{9.14}+\dfrac{3}{14.19}+......................+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)

\(A.\dfrac{5}{3}=\dfrac{5}{9.14}+\dfrac{5}{14.19}+..................+\dfrac{5}{\left(5n-1\right)\left(5n+1\right)}\)

\(A.\dfrac{5}{3}=\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+..................+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\)

\(A.\dfrac{5}{3}=\dfrac{1}{9}-\dfrac{1}{5n+4}\)

\(A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right):\dfrac{3}{5}\)

\(A=\left(\dfrac{1}{9}-\dfrac{1}{5n+\text{4}}\right).\dfrac{3}{5}\)

\(A=\dfrac{1}{9}.\dfrac{3}{5}-\dfrac{1}{5n+4}.\dfrac{3}{5}\)

\(A=\dfrac{1}{15}-\dfrac{1}{5.\left(5n+4\right)}\)

\(\Rightarrow A< \dfrac{1}{15}\)

\(\Rightarrowđpcm\)

Chúc bn học tốt!!!!!!!!!!

\(\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)

\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)\(=\dfrac{1}{5}\cdot\left(\dfrac{5n+6}{5n+6}-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{5\left(n+1\right)}{5n+6}=\dfrac{n+1}{5n+6}=VP\)

Ta có: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}\)

=\(\dfrac{1}{5}.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)

=\(\dfrac{1}{5}.\left(1-\dfrac{1}{5n+6}\right)\)

= \(\dfrac{1}{5}.\left(\dfrac{5n+6}{5n+6}-\dfrac{1}{5n+6}\right)\)

=\(\dfrac{1}{5}.\dfrac{5n+5}{5n+6}\)

=\(\dfrac{1}{5}.\dfrac{5.\left(n+1\right)}{5n+6}\)

=\(\dfrac{n+1}{5n+6}\left(ĐPCM\right)\)

\(VT=\dfrac{3}{9\cdot14}+\dfrac{3}{14\cdot19}+...+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)

\(=\dfrac{3}{5}\left(\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right)\)\(=\dfrac{3}{5}\left(\dfrac{5n+4}{9\left(5n+4\right)}-\dfrac{9}{9\left(5n+4\right)}\right)\)

\(=\dfrac{3}{5}\cdot\dfrac{5\left(n-1\right)}{9\left(5n+4\right)}\)\(=\dfrac{n-1}{3\left(5n+4\right)}< \dfrac{1}{15}=VP\forall n\in N;n\ge2\)

\(\frac{A}{n}=\frac{4n+4}{n}=4+\frac{4}{n}\)
\(\Rightarrow n\in U\left(4\right)\)
Lập bảng tiếp nhé!
\(\frac{B}{n}=\frac{5n+6}{n}=5+\frac{6}{n}\)
Lập bảng

\(2.\)
a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=\frac{3}{29}\cdot\frac{29}{3}-\frac{1}{5}\cdot\frac{29}{3}=1-\left(1+\frac{14}{15}\right)=1-1-\frac{14}{15}=\frac{14}{15}\)
b)\(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}=\frac{5}{9}\cdot\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)

Gọi ƯCLN(4n+3,5n+4)=d(d\(\inℕ^∗\))

\(\Rightarrow\)\(\hept{\begin{cases}4n+3⋮d\\5n+4⋮d\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}5\left(4n+3\right)⋮d\\4\left(5n+4\right)⋮d\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}20n+15⋮d\\20n+16⋮d\end{cases}}\)\(\Rightarrow\)(20n+16-20n-15)\(⋮\)d \(\Rightarrow\)1\(⋮\)d \(\Rightarrow\)d=1

\(\Rightarrow\)ƯCLN(4n+3,5n+4)=1 \(\Rightarrow\)Phân số \(\frac{4n+3}{5n+4}\)là phân số tối giản

Vậy phân số \(\frac{4n+3}{5n+4}\)là phân số tối giản (đpcm)