a) Để A là một phân số thì mẫu của \(A\ne0\) hay \(2n+3\ne0\)

\(\Leftrightarrow n\ne\dfrac{-3}{2}\)

b) Ta có : \(A=\dfrac{12n+1}{2n+3}\)

\(\Rightarrow A=\dfrac{12n+18-17}{2n+3}=\dfrac{12n+18}{2n+3}-\dfrac{17}{2n+3}\)

\(\Rightarrow A=\dfrac{6\left(2n+3\right)}{2n+3}-\dfrac{17}{2n+3}=6-\dfrac{17}{2n+3}\)

Để \(A\in Z\Leftrightarrow\dfrac{17}{2n+3}\in Z\)

\(\Leftrightarrow2n+3\in U\left(17\right)\)

mà \(U\left(17\right)=\left(1;-1;17;-17\right)\)

\(\Rightarrow n\in\left(-1;-2;7;-10\right)\) 

Vậy \(A\in Z\Leftrightarrow n\in\left(-1;-2;7;-10\right)\)

Số học sinh giỏi kì 1 so với cả lớp chiếm số phần là :

 \(\dfrac{3}{3+7}\)  = \(\dfrac{3}{10}\)

Số học sinh giỏi kỳ 2 so với cả lớp chiếm số phần là: 

\(\dfrac{2}{2+3}\) = \(\dfrac{2}{5}\)

Phân số chỉ 4 học sinh là:

\(\dfrac{2}{5}\) - \(\dfrac{3}{10}\) = \(\dfrac{1}{10}\)

Số học sinh cả lớp:

4 : \(\dfrac{1}{10}\) = 40 ( học sinh)

Kết luận số học sinh lớp 6A là : 40 học sinh 

a, \(-\dfrac{315}{540}\) = \(\dfrac{-315:45}{540:45}\) = \(\dfrac{-7}{12}\)        b,   \(\dfrac{25.13}{26.35}\) = \(\dfrac{25.13:5:13}{26.35:13:5}\) = \(\dfrac{5}{14}\)

c,     \(\dfrac{6.9-2.17}{63.3-119}\) = \(\dfrac{2.3.9-2.17}{7.9.3-7.17}\) = \(\dfrac{2.(27-17)}{7.(7-17)}\) = \(\dfrac{2}{7}\)

d, \(\dfrac{3.13-13.18}{15.40-80}\) = \(\dfrac{3.13(1-6)}{40.(15-2)}\) = \(\dfrac{-3.13.5}{40.13}\) = \(\dfrac{-15}{40}\) = \(\dfrac{-15:5}{40:5}\) = \(-\dfrac{3}{8}\)

\(\dfrac{28}{5}\)x - \(\dfrac{23}{6}\)= \(\dfrac{1}{6}\)

\(\dfrac{28}{5}\)x        = \(\dfrac{1}{6}\) + \(\dfrac{23}{6}\)

\(\dfrac{28}{5}\)x        = 4

     x        = 4 : \(\dfrac{28}{5}\)

    x        =   \(\dfrac{5}{7}\)

\(-\dfrac{5}{2}\) <  \(\dfrac{3}{x}\) < \(\dfrac{2}{-3}\)

\(-\dfrac{30}{12}\) < \(\dfrac{30}{10x}\) < \(\dfrac{30}{-45}\)

-45 < 10x < -12

- \(\dfrac{9}{2}\) < x < - \(\dfrac{6}{5}\)

-4,5 < x < - 1,2

x ϵ { -4; -3; -2 }

\(\dfrac{4}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{2}{5}\cdot\dfrac{-3}{7}=\dfrac{15}{4}\)

\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{-6}{35}=\dfrac{15}{4}\)

\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)+\dfrac{6}{35}=\dfrac{15}{4}\)

\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)=\dfrac{15}{4}-\dfrac{6}{35}=\dfrac{501}{140}\)

\(\Rightarrow\dfrac{2}{3}x-\dfrac{9}{5}=\dfrac{501}{140}:\dfrac{2}{5}=\dfrac{501\cdot5}{28\cdot5\cdot2}=\dfrac{501}{56}\)

\(\Rightarrow\dfrac{2}{3}x=\dfrac{501}{56}+\dfrac{9}{5}=\dfrac{2001}{280}\)

\(\Rightarrow x=\dfrac{2001}{280}:\dfrac{2}{3}=\dfrac{2001\cdot3}{280\cdot2}=\dfrac{6003}{560}\)

Ta dùng thuật toán Euclid

1024:580 dư 444

=> ƯCLN(1024; 580)=ƯCLN(580; 444)

580:444 dư 136

=> ƯCLN(580; 444)=ƯCLN(444; 136)

444:136 dư 36

=> ƯCLN(444; 136)=ƯCLN(136; 36)

136:36 dư 28

=> ƯCLN(136; 36)=ƯCLN(36;28)

Ta có: 36=2².3²

28=2².7

=> ƯCLN(36; 28)=ƯCLN(1024; 580)=2²=4

Làm theo cách làm tương tự, ta có ƯCLN(690; 960)=60

\(\dfrac{-5}{12}.\dfrac{2}{11}+\dfrac{-5}{12}.\dfrac{9}{11}+\dfrac{5}{12}\)

\(=\dfrac{-5}{12}\left(\dfrac{2}{11}+\dfrac{9}{11}-1\right)\)

\(=\dfrac{-5}{12}\left(1-1\right)\)

\(=0\)

\(=\left(-67\right).\left(-33\right)-67.77+\left(-67\right).33-77.33\)

\(=\left[\left(-67\right).\left(-33\right)+\left(-67\right).33\right]-66.77-77.33\)

\(=-77\left(66+33\right)=-77.99=-7623\)