Question

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Answer 1

[Lê Song Phương]

Câu 15: 

Ta có \(x^2-4x+y^2-6y+15=2\Leftrightarrow x^2-4x+y^2-6y+13=0\) \(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-6y+9\right)=0\) \(\Leftrightarrow\left(x-2\right)^2+\left(y-3\right)^2=0\) (*)

Vì \(\left(x-2\right)^2\ge0;\left(y-3\right)^2\ge0\) nên (*) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy cặp số \(\left(x;y\right)\) thỏa mãn pt đã cho là \(\left(2;3\right)\)

Answer 2

[Đỗ Tuấn Phong]

 

$x^2-4x+y^2-6y+15=2\iff x^2-4x+y^2-6y+13=0$ $\iff \left(x^2-4x+4\right)+\left(y^2-6y+9\right)=0$ $\iff {\left(x-2\right)}^2+{\left(y-3\right)}^2=0$ (*)

Vì ${\left(x-2\right)}^2\ge 0;{\left(y-3\right)}^2\ge 0$ => (*) xảy ra $\iff \{\begin{array}{c}{\left(x-2\right)}^2=0\\ {\left(y-3\right)}^2=0\end{array}\iff \{\begin{array}{c}x-2=0\\ y-3=0\end{array}\iff \{\begin{array}{c}x=2\\ y=3\end{array}$