Question

[Phan Ngọc ánh]

 Tìm x , y :

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

Biết x < y 

x = ....... , y = ........

x = ....... , y = ........

x = ....... , y = ........

Answer 1

[Tạ Đức Hoàng Anh]

- Ta có: \(\frac{3}{x}+\frac{y}{3}=\frac{5}{6}\)

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

       \(\Leftrightarrow\frac{3}{x}=\frac{5-2y}{6}\)

      \(\Leftrightarrow x.\left(5-2y\right)=18=1.18=\left(-1\right).\left(-18\right)=2.9=\left(-2\right).\left(-9\right)=3.6=\left(-3\right).\left(-6\right)\)

- Vì \(5-2y\)là lẻ \(\Rightarrow\)\(5-2y\in\left\{-9,-3,-1,1,3,9,\right\}\)

- Ta có bảng giá trị:

\(5-2y\)	\(-9\)	\(-3\)	\(-1\)	\(1\)	\(3\)	\(9\)
\(x\)	\(-2\)	\(-6\)	\(-18\)	\(18\)	\(6\)	\(2\)
\(y\)	\(7\)	\(4\)	\(3\)	\(2\)	\(1\)	\(-2\)
	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(TM\right)\)	\(\left(L\right)\)	\(\left(L\right)\)	\(\left(L\right)\)

Vậy \(\left(x,y\right)\in\left\{\left(-2,7\right);\left(-6,4\right);\left(-18,3\right)\right\}\)

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