Question

[Nguyễn Ngọc Gia Hân]

Tìm x ; y \(\in\) N :

a. \(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

b. 3x - xy -4y + 12 = 17

Answer 1

[Nguyễn Thanh Hằng]

a) Ta có :

\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{4}{x}\)

\(\Leftrightarrow\dfrac{5}{6}-\dfrac{2y}{6}=\dfrac{4}{x}\)

\(\Leftrightarrow\dfrac{5-2y}{6}=\dfrac{4}{x}\)

\(\Leftrightarrow\left(5-2y\right)x=6.4=24\)

Vì \(x,y\in N\Leftrightarrow5-2y\in N;5-2y;x\inƯ\left(24\right)\)

Ta có bảng :

\(x\)	\(y\)	\(5-2y\)	\(Đk\) \(x,y\in N\)
\(1\)	\(\dfrac{-19}{2}\)	\(24\)	loại
\(2\)	\(\dfrac{-7}{2}\)	\(12\)	loại
\(3\)	\(\dfrac{-3}{2}\)	2\(8\)	loại
\(4\)	\(\dfrac{1}{2}\)	\(6\)	loại
\(8\)	\(1\)	\(3\)	thỏa mãn
\(12\)	\(\dfrac{3}{2}\)	\(2\)	loại
\(24\)	\(2\)	\(1\)	thỏa mãn

Vậy ...

Answer 2

[Mashiro Shiina]

\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{2y}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{5-2y}{6}\)

\(\Rightarrow x\left(5-2y\right)=24\)

\(\Rightarrow x;5-2y\inƯ\left(24\right)\)

Xét ước là xong

\(3x-xy-4y+12=17\)

\(\Rightarrow x\left(3-y\right)+4\left(3-y\right)=17\)

\(\Rightarrow\left(x+4\right)\left(3-y\right)=17\)

\(\Rightarrow x+4;3-y\inƯ\left(17\right)\)

\(Ư\left(17\right)=\left\{\pm1;\pm17\right\}\)

Xét ước