Tìm số tự nhiên a biết: a, 10 chia hết cho 3a cộng 1 b, acộng 6

a, 10⋮3a+1 => 3a+1∈Ư(10) => 3a+1∈{1;2;5;10} => a∈{0;13;43;3}. Vì a∈N, a∈{0;3}

b, a+6⋮a+1 => a+1+5⋮a+1 => 5⋮a+1 => a+1∈Ư(5) =>  a+1∈{1;5} => a∈{0;4}

c, 3a+7⋮2a+3 => 2.(3a+7)-3(2a+3)⋮2a+3 => 5⋮2a+3 => 2a+3∈Ư(5)

=> 2a+3∈{1;5} => a = 1

d, 6a+11⋮2a+3 => 3.(2a+3)+2⋮2a+3 => 2⋮2a+3 => 2a+3∈Ư(2)

=> 2a+3∈{1;2} => a∈∅