Question

Tìm hai số hữu tỉ \(x\) và \(y\) sao cho \(x-y=x.y=x:y,\left(y\ne0\right)\)

Answer 1

ta có:

\(x+y=x.y\)

\(\Rightarrow y=x.y-x=x.(y-1)\)

\(\Rightarrow x:y=y-1=x+y\)

\(\Rightarrow x=-1\)

\(thay\) \(x+y=x.y\)

\(\Rightarrow y-1=-y\Rightarrow2y=1\Rightarrow y=\dfrac{1}{2}\)

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

Answer 2

\(\dfrac{1}{y}=\dfrac{x}{4}-\dfrac{1}{2}=\dfrac{x-2}{4}=>y.\left(x-2\right)=4\)

Vì x ,y \(\in\) z nên x - 2 \(\in\) z , ta có bảng sau :

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

Answer 3

\(x-y=x.y\)

=> \(x=x.y+y=y.\left(x+1\right)\)

\(x:y=y.\left(x+1\right):y=x+1\)

=> \(x-y=x+1=>y=-1\)

\(x=\left(-1\right)\left(x+1\right)=>x=-x-1\)

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

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

Answer 4

x – y = x.y ⇒ x = x.y + y = y.(x + 1)

x : y = y.(x + 1) : y = x + 1

⇒ x – y = x + 1 ⇒ y = −1

x = (-1)(x + 1) ⇒ x = − x – 1 ⇒ 2x = −1 ⇒ x = (-1)/2

Vậy x = −12;y = −1;x = −12;y = −1