Giải hệ phương trình \(\left\{ {\begin{array}{*{20...

B

- Hướng dẫn giải

Điều kiện: \({y^2} \ge {x^2}\)

\(\begin{array}{l}
\left\{ {\begin{array}{*{20}{c}}
{x + \sqrt {{y^2} - {x^2}}  = 12 - y\,\,\,(1)}\\
{x\sqrt {{y^2} - {x^2}}  = 12\,\,\,\,\,\,\,\,\,\,\,\,\,(2)}
\end{array}} \right.\\
(1) \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{12 - y \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{{x^2} + 2x\sqrt {{y^2} - {x^2}}  + {y^2} - {x^2} = 144 - 24y + {y^2}}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{y \le 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{2x\sqrt {{y^2} - {x^2}}  = 144 - 24y\,(*)}
\end{array}} \right.} \right.
\end{array}\)

Thế (2) vào (*) ta được: \(2.12 = 144 - 24y \Leftrightarrow 24y = 120 \Leftrightarrow y = 5\,\,(tm)\)

\(\begin{array}{l}
 =  > x\sqrt {25 - {x^2}}  = 12 \Leftrightarrow {x^2}\left( {25 - {x^2}} \right) = 144\\
 \Leftrightarrow {x^4} - 25{x^2} + 144 = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{{x^2} = 16}\\
{{x^2} = 9}
\end{array}} \right.\\
 \Rightarrow T = x_1^2 + x_2^2 - y_1^2 = 16 + 9 - {5^2} = 0
\end{array}\)