d:
ĐKXĐ: y<>0; x<>0; y<>2
\(\dfrac{4}{x}+\dfrac{2}{y}=1\)
=>\(\dfrac{4y}{xy}+\dfrac{2x}{xy}=1\)
=>2x+4y=xy
=>x(2-y)=-4y
=>x(y-2)=4y
=>\(x=\dfrac{4y}{y-2}\)
mà x,y nguyên
nên \(4y⋮y-2\)
\(\Leftrightarrow4y-8+8⋮y-2\)
=>\(y-2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>\(y\in\left\{3;1;4;6;-2;10;-6\right\}\)
=>\(x\in\left\{12;-4;8;6;2;5;3\right\}\)
e:
ĐKXĐ: x<>0; y<>0; y<>3
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3}\)
=>\(\dfrac{x+y}{xy}=\dfrac{1}{3}\)
=>3x+3y=xy
=>x(3-y)=-3y
=>\(x=\dfrac{3y}{y-3}\)
mà x,y nguyên
nên \(3y⋮y-3\)
=>\(3y-9+9⋮y-3\)
=>\(y-3\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(y\in\left\{4;2;6;12;-6\right\}\)
=>\(x\in\left\{12;-6;6;4;2\right\}\)
