Question

[Le quy mui]

Tìm các cặp số nguyên x,y biết

a,\(2x^2+y^2+6=4\left(x-y\right)\)

b,\(x^2\left(y+2\right)+1=y^2\)

Answer 1

[Cold Wind]

a) \(2x^2+y^2+6=4\left(x-y\right)\)

\(\Leftrightarrow2x^2+y^2+6-4x+4y=0\)

\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(y^2+4y+4\right)=0\)

\(\Leftrightarrow2\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Answer 2

[alibaba nguyễn]

b/ x²(y + 2) + 1 = y²

<=> x²(y + 2) + 1 = (y + 2)(y - 2) + 4

<=> (y + 2)(x² + 2 - y) = 3

Làm tiếp nhé