$a) (2x+1).y=4x+1 \rightarrow y=\dfrac{4x+1}{2x+1}$
Vì y nguyên nên $\dfrac{4x+1}{2x+1}\in Z$
$\rightarrow 4x+1 \vdots 2x+1$
$\rightarrow 2(2x+1)-1 \vdots 2x+1$
$\rightarrow 1 \vdots 2x+1$
$\rightarrow 2x+1 \in \{ 1;-1 \}$
$\rightarrow 2x \in \{0;-2 \}$
$\rightarrow x\in \{ 0; -1 \}$
Khi x=0 thì y=1
Khi x=-1 thì y=3
$b)(x-1).y=x^2+7 \rightarrow y=\dfrac{x^2+7}{x-1}$
Vì y nguyên nên $\dfrac{x^2+7}{x-1} \in Z$
$\rightarrow x^2+7 \vdots x-1$
$\rightarrow x^2-x+x+7 \vdots x-1$
$\rightarrow x(x-1)+x+7 \vdots x-1$
$\rightarrow x+7 \vdots x-1$
$\rightarrow x-1+8 \vdots x-1$
$\rightarrow 8 \vdots x-1$
$\rightarrow x-1\in \{1;2;4;8;-1;-2;-4;-8 \}$
$\rightarrow x\in \{ 2;3;5;9;0;-1;-3;-7 \}$
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$c)(2x+1).y=x^2-3 \rightarrow y=\dfrac{x^2-3}{2x+1}$
Vì y nguyên nên $\dfrac{x^2-3}{2x+1} \in Z$
$\rightarrow x^2-3 \vdots 2x+1$
$\rightarrow 2x^2-6 \vdots 2x+1$
$\rightarrow 2x^2+x-x-3 \vdots 2x+1$
$\rightarrow x(2x+1)-(x+3) \vdots 2x+1$
$\rightarrow x+3 \vdots 2x+1$
$\rightarrow 2x+6 \vdots 2x+1$
$\rightarrow 2x+1+5 \vdots 2x+1$
$\rightarrow 5 \vdots 2x+1$
$\rightarrow 2x+1\in \{1;5;-1;-5 \}$
$\rightarrow 2x\in \{0;4;-2;-6 \}$
$\rightarrow x\in \{0;2;-1;-3 \}$
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