4x+57 \(⋮\)2x+1
Ta có: 2x+1\(⋮\)2x+1
=>2.(2x+1)\(⋮\)2x+1
=>4x+2\(⋮\)2x+1(1)
Theo bài ta có:4x+57\(⋮\)2x+1(2)
Từ (1) và(2) suy ra (4x+57)-(4x+2)\(⋮\)2x+1
=>4x+57-4x-2\(⋮\)2x+1
=>55\(⋮\)2x+1
=>2x+1\(\in\)Ư(55)={1;5;11;55}
+)2x+1=1=>2x=1-1=>2x=0=>x=0:2=>x=0
+2x+1=5=>2x=5-1=>2x=4=>x=4:2=>x=2
+)2x+1=11=>2x=11-1=>2x=10=>x=10:2=>x=5
+)2x+1=55=>2x=55-1=>2x=54=>x=54:2=>x=27
Vậy x\(\in\){0;2;5;27}
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