Có 2x +7= 2x-4+11
=> 2x-4\(⋮\)x-4 =>2x+7\(⋮\)x-4 => 11\(⋮\)x-4 <=> x-4 \(\in\)Ư(11)
Mà Ư(11)=1,-1,11,-11 \(\Rightarrow\)x-4\(\in\)1,-1,-11,11 \(\Rightarrow\)x\(\in\)5,3,-7,15
2x + 7 \(⋮\)x - 4
x - 4 \(⋮\)x - 4 => 2( x-4 ) \(⋮\)x - 4 => 2x - 8 \(⋮\)x - 4
=> ( 2x + 7 ) - ( 2x - 8 ) \(⋮\)x - 4
15 \(⋮\)x - 4
=> x - 4 \(\in\){ 1 ; 15 ; - 1 ; -15 ; 3 ; 5 ; - 3 ; - 5 }
x \(\in\){ 5 ; 19 ; 3 ; -11 ; 7 ; 9 ; 1 ; -1 }
Xét \(2x+7\)
\(\Rightarrow2x+7=2x-8+15\)
Ta có phân số :
\(\frac{2x-8+15}{x-4}\)
\(=2+\frac{15}{x-4}\)
Vậy để \(2x+7⋮x-4\)
\(\Rightarrow15⋮x-4\)
\(\Rightarrow x-4\inƯ\left(15\right)=\left(1;-1;3;-3;-5;5;-15;15\right)\)
\(\Rightarrow x\in\left(5;4;1;-1;9;-11;19\right)\)
2x + 7 \(⋮\)x - 4
x - 4 \(⋮\)x - 4
2(x - 4) \(⋮\)x - 4
2x - 8 \(⋮\)x - 4
\(\Rightarrow\left[\left(2x+7\right)-\left(2x-8\right)\right]⋮x-4\)
\(\left(2x+7-2x+8\right)⋮x-4\)
\(15⋮x-4\)
\(x-4\in U\left(15\right)=\left\{1;-1;5;-5;15;-15;3;-3\right\}\)
x - 4 = 1 => x = 4+1 = 5
x - 4 = -1 => -1 + 4 = 3
x- 4 = 5 => 5 + 4 = 9
x - 4 = -5 => -5 + 4 = -1
x - 4 = 3 => 3 + 4 = 7
x - 4 = -3 => -3 + 4 = 1
x - 4 = 15 => 15 + 4 = 19
x - 4 = -15 => -15 + 4 = -11
Vậy x \(\in\){5;3;9;-1;1;7;19;11}