Question

Tìm số nguyên x ; y biết rằng  \(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)

Answer 1

=> \(\frac{xy-4}{4y}=\frac{1}{2}\) => 2(xy-4) = 1.4y => 2xy - 8 = 4y => 2xy - 4y = 8 => 2y(x - 2) = 8 => y(x -2) = 4

x;y nguyên nên y \(\in\) Ư(4) = {-4;-2;-1;1;2;4}

Tương ứng x - 2 \(\in\) {-1;-2;-4;4;2;1} => x \(\in\) {1;0;-2;6;4;3}

Vậy...

Answer 2

Ta có: \(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)

=>\(\frac{1}{y}=\frac{x}{4}-\frac{1}{2}\)

=>\(\frac{1}{y}=\frac{x-2}{4}\)

=>

y.(x-2)=4

Ta thấy: 4=1.4=2.3=(-1).(-4)=(-2).(-2)

x-2	1	4	-1	-4	2	-2
x	3	6	1	-2	4	0
y	4	1	-4	-1	2	-2

          Vậy (x,y)=(3,4),(6,1),(1,-4),(-2,-1),(4,2),(0,-2)

Answer 3

\(\frac{x}{4}-\frac{1}{y}=\frac{1}{2}\)

=> \(\frac{x}{4}-\frac{1}{2}=\frac{1}{y}\)

=> \(\frac{x}{4}-\frac{2}{4}=\frac{1}{y}\)

=> \(\frac{x-2}{4}=\frac{1}{y}\)

=> y.(x-2)= 4

Vì x;y \(\in\)Z

=> x-2; y \(\in\)Z

=> ( x-2;y)=(-1;-4);(1;4);(-2;-2);(2;2);(-4;-1);(4;1)

=> (x;y)=(1;-4);(3;4);(0;-2);(4;2);((-2;-1);(6;1)