bài 1:
b) 4n+3 \(⋮\)11
\(\Rightarrow\)4n+3+11\(⋮\)11
\(\Rightarrow\)4n+14\(⋮\)11
\(\Rightarrow\)2.(2n+7)\(⋮\)11
\(\Rightarrow\)2n+7\(⋮\)11
\(\Rightarrow\)2n+7+11\(⋮\)11
\(\Rightarrow\)2n+18\(⋮\)11
\(\Rightarrow\)2.(n+9)\(⋮\)11
\(\Rightarrow\)n+9\(⋮\)11
\(\Rightarrow\)n=11k+2
\(\Rightarrow\)n\(\ne\)11k+2
4n+3\(⋮\)17
\(\Rightarrow\)3n+3+17\(⋮\)17
\(\Rightarrow\)4n+20\(⋮\)17
\(\Rightarrow\)4.(n+5)\(⋮\)17
\(\Rightarrow\)n+5\(⋮\)17
\(\Rightarrow\)n=17k+12
\(\Rightarrow\)n\(\ne\)17k+12
Vậy n\(\ne\)17k+12 và n\(\ne\)11k+2 (k\(\in\)N) thì A là tối giản