Question

Bài 1 : Tìm các cặp số nguyên ( x;y ) \(\in Z\) biết

\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

Bài 2 : Tìm các cặp số nguyên ( x;y ) \(\in Z\) biết \(x\left(x+y\right)=\dfrac{1}{48};y\left(x+y\right)=\dfrac{1}{24}\)

Answer 1

Answer 2

\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)

\(\dfrac{1}{x}=\dfrac{1+2y}{6}\)

\(6=x\left(1+2y\right)\)

Tự làm típ

\(x\left(x+y\right)=\dfrac{1}{48};y\left(x+y\right)=\dfrac{1}{24}\)

\(x^2+xy=\dfrac{1}{48};xy+y^2=\dfrac{1}{24}\)

\(\Rightarrow x^2+xy-y^2-xy=\dfrac{1}{48}-\dfrac{1}{24}\)

\(x^2-y^2=\dfrac{-1}{24}\)

\(\left(x+y\right)\left(x-y\right)=\dfrac{-1}{24}\)(HĐT số 3)

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