x^2+y^2-xy-3x-3y+9=0 Question x^2+y^2-xy-3x-3y+9=0 in progress 0 Môn Toán Ben Gia 3 years 2021-05-24T18:42:31+00:00 2021-05-24T18:42:31+00:00 1 Answers 19 views 0
Answers ( )
Đáp án:
$\left( {x;y} \right) = \left( {3;3} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{x^2} + {y^2} – xy – 3x – 3y + 9 = 0\\
\Leftrightarrow 2{x^2} + 2{y^2} – 2xy – 6x – 6y + 18 = 0\\
\Leftrightarrow \left( {{x^2} – 2xy + {y^2}} \right) + \left( {{x^2} – 6x + 9} \right) + \left( {{y^2} – 6y + 9} \right) = 0\\
\Leftrightarrow {\left( {x – y} \right)^2} + {\left( {x – 3} \right)^2} + {\left( {y – 3} \right)^2} = 0\\
\Leftrightarrow {\left( {x – y} \right)^2} = {\left( {x – 3} \right)^2} = {\left( {y – 3} \right)^2} = 0\\
\left( {do:{{\left( {x – y} \right)}^2} + {{\left( {x – 3} \right)}^2} + {{\left( {y – 3} \right)}^2} \ge 0,\forall x,y,z} \right)\\
\Leftrightarrow x – y = x – 3 = y – 3 = 0\\
\Leftrightarrow x = y = 3
\end{array}$
Vậy $\left( {x;y} \right) = \left( {3;3} \right)$