Toán Lớp 10: 3) √x+3 + √6-x= 3 + √(x+3)(6-x)
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Toán Lớp 10: 3) √x+3 + √6-x= 3 + √(x+3)(6-x)
Comments ( 1 )
x = – 3\\
x = 6
\end{array} \right.\)
DK:6 \ge x \ge – 3\\
\sqrt {x + 3} + \sqrt {6 – x} = 3 + \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} \\
Đặt:\sqrt {x + 3} + \sqrt {6 – x} = t\left( {t \ge 0} \right)\\
\to x + 3 + 2\sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} + 6 – x = {t^2}\\
\to 9 + 2\sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = {t^2}\\
\to \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = \dfrac{{{t^2} – 9}}{2}\\
Pt \to t = 3 + \dfrac{{{t^2} – 9}}{2}\\
\to {t^2} – 9 + 6 = 2t\\
\to {t^2} – 2t – 3 = 0\\
\to \left( {t – 3} \right)\left( {t + 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = 3\\
t = – 1\left( l \right)
\end{array} \right.\\
\to 9 + 2\sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = 9\\
\to \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = 0\\
\to \left[ \begin{array}{l}
x = – 3\\
x = 6
\end{array} \right.
\end{array}\)