\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)

\(=\frac{100}{200}=\frac{1}{2}\)

\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

\(< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)

\(=\frac{100}{100}=1\)

Suy ra dpcm. 

ta có:\(\frac{n-7}{n+1}=\frac{n+1-8}{n+1}=1-\frac{8}{n+1}\)

để \(n-7⋮n+1\Rightarrow\frac{n-7}{n+1}\in Z\)

\(\Rightarrow1-\frac{8}{n+1}\in Z\Leftrightarrow\frac{8}{n+1}\in Z\)

\(\Rightarrow n+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

do n nguyên dương nên \(\Rightarrow n+1\in\left\{1;2;4;8\right\}\)

bạn tính nốt n nhé

Ta có :

3n+5 \(⋮\)n-4

Mà 3(n-4) hay 3n-12 \(⋮\)n-4

\(\Rightarrow\left(3n+5\right)-\left(3n-12\right)⋮n-4\)

\(17⋮n-4\)

\(\Rightarrow n-4\inƯ\left(17\right)\)

Còn lại tự lm nha.

HT

ta có:\(B=\frac{10n-3}{4n-10}=\frac{5.\left(2n-5\right)+22}{2.\left(2n-5\right)}=\frac{5}{2}+\frac{22}{2.\left(2n-5\right)}=\frac{5}{2}+\frac{11}{2n-5}\)

\(Bmax\Leftrightarrow\frac{5}{2}+\frac{11}{2n-5}max\Leftrightarrow\frac{11}{2n-5}max\Leftrightarrow2n-5=1\)

\(\Leftrightarrow2n=6\Leftrightarrow n=3\)

\(B=\frac{5}{2}+11=\frac{27}{2}\)

VẬY \(n=3\) THÌ \(maxB=\frac{27}{2}\)