Ta có:

Để \(\dfrac{n+4}{3n+5}\) đạt giá trị nguyên thì \(\left(n+4\right)⋮\left(3n+5\right)\)

\(\Rightarrow3\left(n+4\right)⋮3n+5\)

\(\Rightarrow\left(3n+5+7\right)⋮\left(3n+5\right)\)

\(\Rightarrow7⋮\left(3n+5\right)\)

\(\Rightarrow3n+5\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Rightarrow3n\in\left\{-12;-6;-4;2\right\}\)

\(\Rightarrow n\in\left\{-4;-2;-\dfrac{4}{3};\dfrac{2}{3}\right\}\)

a) \(\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

\(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)

\(=\left(6+3-4\right)+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)+\dfrac{7}{11}\)

\(=5+\dfrac{7}{11}\)

\(=\dfrac{62}{11}\)

b) \(-1\dfrac{1}{3}-2\dfrac{5}{6}.\dfrac{6}{11}+3:5\%\)

\(=-\dfrac{4}{3}-\dfrac{17}{6}.\dfrac{6}{11}+3:\dfrac{1}{20}\)

\(=-\dfrac{4}{3}-\dfrac{17}{11}+60\)

\(=-\dfrac{95}{33}+60\)

\(=\dfrac{1885}{33}\)

c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

\(=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{99-97}{97.99}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

d) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

Em bổ sung cho đầy đủ đề nhé

Số các số chia 4 dư 2:

(30 - 2) : 4 + 1 = 8 (số)

Xác suất của biến cố "Số xuất hiện trên thẻ được rút ra là số chia 4 dư 2":

8/30 = 4/15

12/5 - 3/5 - 5/8

= 12/5 - 1

= 7/5

Do 38,1 < x < 39,1 nên x = 39