Ta có: \(\frac{n+5}{n+2}\)=\(\frac{n+2+3}{n+2}\)= \(\frac{3}{n+2}+1\)

Để \(\frac{n+5}{n+2}\)\(\inℤ\)khi x\(\inℤ\)thì

3\(⋮\)n+2 => n+2\(\inƯ\left(3\right)\)=(1,-1,3,-3)

=> n\(\in\)(-1;-3;1;-5)

bạn ơi ở dòng cuối cùng nó là n thuộc {-1;-3;1;3} nha bạn

Để \(\frac{5}{n}\)đạt giá trị lớn nhất 

=> n nhỏ nhất 

mà n \(\inℕ\Rightarrow n=1\)

Khi đó \(\frac{5}{n}=\frac{5}{1}=5\)

Vậy \(\frac{5}{n}\)có giá trị lớn nhất là 5

C= 5/2 . (1/2-1/4+1/4-1/6+1/6-1/8+......+1/2020-1/2022)

C= 5/2 . (1/2-1/20220)

C= 5/2. 505/1011

c=2525/2022

\(C=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{2020.2022}\)

\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)

\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)=\frac{5}{2}.\frac{1010}{2022}=\frac{5050}{4044}=\frac{2525}{2022}\)

\(C=\frac{5}{2\cdot4}+\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+...+\frac{5}{2020\cdot2022}\)

\(C=5\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{2020\cdot2022}\right)\)

\(C=\frac{5}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2020\cdot2022}\right)\)

\(C=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)

\(C=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{2022}\right)\)

\(C=\frac{5}{2}\cdot\frac{505}{1011}\)

\(C=\frac{2525}{2022}\)

số hs giỏi là

    48.1/6=8(hs)

số hs trung bình là

    8.300%=24(hs)

số hs khá là

     48-8-24=16(hs)

          Đ/S................