Câu hỏi của Uchiha Itachi

Question

Tìm x, y ∈ Z thoả mãn: 12x² + 6xy + 3y² = 28(x + y)

Nguyễn Việt Lâm · Accepted Answer

$\Leftrightarrow9x^2+3\left(x^2+2xy+y^2\right)=28\left(x+y\right)$

$\Leftrightarrow9x^2+3\left(x+y\right)^2=28\left(x+y\right)$

$\Leftrightarrow9x^2=-3\left(x+y\right)^2+28\left(x+y\right)$

$\Leftrightarrow9x^2=-3\left(x+y-\frac{14}{3}\right)^2+\frac{196}{3}\le\frac{196}{3}$

$\Rightarrow x^2\le7\Rightarrow x^2=\left\{0;1;4\right\}\Rightarrow x=\left\{-2;-1;0;1;2\right\}$

Thế vào pt ban đầu để tìm y nguyên

Câu trả lời:

$\Leftrightarrow9x^2+3\left(x^2+2xy+y^2\right)=28\left(x+y\right)$

$\Leftrightarrow9x^2+3\left(x+y\right)^2=28\left(x+y\right)$

$\Leftrightarrow9x^2=-3\left(x+y\right)^2+28\left(x+y\right)$

$\Leftrightarrow9x^2=-3\left(x+y-\frac{14}{3}\right)^2+\frac{196}{3}\le\frac{196}{3}$

$\Rightarrow x^2\le7\Rightarrow x^2=\left\{0;1;4\right\}\Rightarrow x=\left\{-2;-1;0;1;2\right\}$

Thế vào pt ban đầu để tìm y nguyên