Question

tìm số nguyên x,y biết:

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{2}\)

Answer 1

`1/x-1/y=2`
`đk:x,y ne 0`
Nhân 2 vế với `xy ne 0` ta có:
`y-x=2xy`
`=>2y-2x=4xy`
`=>4xy-2x=2y`
`=>2x(2y-1)=2y-1+1`
`=>(2y-1)(2x-1)=1`
Vì `x,y ne 0=>2x-1,2y-1 ne -1`
`=>2x-1=2y-1=1`
`=>x=y=1`

Answer 2

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{2}\Leftrightarrow\dfrac{x-y}{xy}=\dfrac{1}{2}\)

do x,y nguyên nên:\(\left\{{}\begin{matrix}x-1=1\\x.y=2\end{matrix}\right.\)

ta có :2=1.2

mã-y=1⇒x>y⇒x=2,y=1

Answer 3

Giải:

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{2}\) 

\(\Rightarrow\dfrac{\left(y-x\right)}{\left(xy\right)}=\dfrac{1}{2}\) 

\(\Rightarrow2.\left(y-x\right)=xy\) 

\(\Rightarrow2y-xy-2x=0\) 

\(\Rightarrow y.\left(2-x\right)+2.\left(2-x\right)=4\) 

\(\Rightarrow\left(y+2\right).\left(2-x\right)=4\) 

\(\Rightarrow\left(y+2\right)\) và \(\left(2-x\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)  

Ta có bảng giá trị:

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

Vậy \(\left(x;y\right)=\left\{\left(3;-6\right);\left(4;-4\right);\left(6;-3\right);\left(-2;-1\right);\left(0;0\right);\left(1;2\right)\right\}\)