Ta có: 4x + 3 = 4x + 2 + 1 = 2( 2x + 1 ) + 1 chia hết cho 2x + 1
=> 1 chia hết cho 2x + 1 => 2x + 1 \(\in\)Ư ( 1 ) = { 1 }
=> 2x + 1 = 1 => 2x = 1 - 1 = 0 => x = 0 : 2 = 0
Vậy ...
Ta có: 4x + 3 = 4x + 2 + 1 = 2( 2x + 1 ) + 1 chia hết cho 2x + 1
=> 1 chia hết cho 2x + 1 => 2x + 1 \(\in\)Ư ( 1 ) = { 1 }
=> 2x + 1 = 1 => 2x = 1 - 1 = 0 => x = 0 : 2 = 0
\(\frac{4x+3}{2x+1}=\frac{2\left(2x+1\right)+1}{2x+1}=\frac{2\left(2x+1\right)}{2x+1}+\frac{1}{2x+1}=2+\frac{1}{2x+1}\in Z\)
=>1 chia hết 2x+1
=>2x+1\(\in\){1;-1}
=>x\(\in\){0} vì x thuộc N