\(\left(10x+33\right)⋮\left(2x+1\right)\)
\(\Rightarrow\left[5\left(2x+1\right)+28\right]⋮\left(2x+1\right)\)
Vì \(\left[5\left(2x+1\right)\right]⋮\left(2x+1\right)\)nên \(28⋮\left(2x+1\right)\)
Mà 2x + 1 là số lẻ nên \(2x+1\in\left\{1;7\right\}\)
\(TH1:2x+1=1\Leftrightarrow x=0\)
\(TH2:2x+1=7\Leftrightarrow x=3\)
\(10x+33⋮2x+1\)
\(\Leftrightarrow5\left(2x+1\right)+28⋮2x+1\)
\(\Leftrightarrow28⋮2x+1\) ( vì \(5\left(2x+1\right)⋮2x+1\))
\(\Leftrightarrow2x+1\inƯ\left(28\right)\)
Mặt khác \(x\in N\Rightarrow2x+1\in N\)và 2x+1 lẻ
\(\Leftrightarrow2x+1\in\left\{1;7\right\}\)
\(\Leftrightarrow x\in\left\{0;3\right\}\)
Cám ơn hai bạn nha !