Question

[phạm tài anh]

tìm x nguyên biết 1/x+1/y=1/5

Answer 1

[Nguyễn Ngọc Huy Toàn]

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\); \(x,y\ne0\)

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{5}-\dfrac{1}{y}\)

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{y-5}{5y}\)

\(\Leftrightarrow x=\dfrac{5y}{y-5}\)

\(\Leftrightarrow x=5+\dfrac{25}{y-5}\)

Để x;y nguyên thì \(y-5\inƯ\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)

`@y-5=1->y=6(tm)` \(\rightarrow x=5+\dfrac{25}{6-5}=30\) (tm)

`@y-5=-1->y=4(tm)` `->x=-20(tm)`

`@y-5=5->y=10(tm)` `->x=10(tm)`

`@y-5=-5->y=0(ktm)` 

`@y-5=25->y=30(tm)` `->x=6(tm)`

`@y-5=-25->y=-20(tm)` `->x=4(tm)`

Vậy ....

Answer 2

[Đoàn Phương Linh]

Ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)

\(\Rightarrow\)\(\dfrac{y}{xy}+\dfrac{x}{xy}=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{x+y}{xy}=\dfrac{1}{5}\)

\(\Rightarrow\)\(5\left(x+y\right)=xy\)

\(\Rightarrow5x+5y=xy\)

\(\Rightarrow5x-xy+5y=0\)

\(\Rightarrow\left(5x-xy\right)+5y=0\)

\(\Rightarrow x\left(5-y\right)+5y=0\)

\(\Rightarrow x\left(5-y\right)-5\left(5-y\right)=0-25\)

\(\Rightarrow\left(5-y\right)\left(x-5\right)=-25\)

Vì x; y \(\in\) Z \(\Rightarrow\)\(\left\{{}\begin{matrix}x-5\in Z\\5-y\in Z\end{matrix}\right.\)

Để \(\left(5-y\right)\left(x-5\right)=-25\)

\(\Rightarrow\)\(5-y\in\)Ư\(_{\left(-25\right)}\)

\(\Rightarrow5-y\in\left\{1;-1;5;-5;25;-25\right\}\)

Ta có bảng:

\(5-y\)	\(1\)	\(-1\)	\(5\)	\(-5\)	\(25\)	\(-25\)
\(x-5\)	\(-25\)	\(25\)	\(-5\)	\(5\)	\(-1\)	\(1\)
\(y\)	\(4\)	\(6\)	\(0\)	\(10\)	\(-20\)	\(30\)
\(x\)	\(-20\)	\(30\)	\(0\)	\(10\)	\(4\)	\(6\)
	chọn	chọn	chọn	chọn	chọn	chọn

vậy \(\left(x;y\right)=\)\(\left(-20;4\right)\)\(;\left(30;6\right)\)\(;\left(0;0\right)\)\(;\left(10;10\right)\)\(;\left(4;-20\right)\)\(;\left(6;30\right)\)