Toán Lớp 10: $\sqrt{3x+1}-$$\sqrt{6-x}=$$-3x^2+14x+8$
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Toán Lớp 10: $\sqrt{3x+1}-$$\sqrt{6-x}=$$-3x^2+14x+8$
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\sqrt {3x + 1} – \sqrt {6 – x} = – 3{x^2} + 14x + 8\\
\Leftrightarrow \sqrt {3x + 1} – \sqrt {6 – x} + 3{x^2} – 14x – 8 = 0\\
\Leftrightarrow \sqrt {3x + 1} – 4 + \left( {\sqrt {6 – x} – 1} \right) + 3{x^2} – 14x – 5 = 0\\
\Leftrightarrow \dfrac{{3x – 15}}{{\sqrt {3x + 1} + 4}} + \dfrac{{x – 5}}{{\sqrt {6 – x} + 1}} + \left( {x – 5} \right)\left( {3x + 1} \right) = 0\\
\Leftrightarrow \left( {x – 5} \right)\left( {\dfrac{3}{{\sqrt {3x + 1} + 4}} + \dfrac{1}{{\sqrt {6 – x} + 1}} + 3x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
\dfrac{3}{{\sqrt {3x + 1} + 4}} + \dfrac{1}{{\sqrt {6 – x} + 1}} + 3x + 1 = 0\left( * \right)
\end{array} \right.\\
x \ge – \dfrac{1}{3} \Rightarrow 3x + 1 \ge 0 \Rightarrow \dfrac{3}{{\sqrt {3x + 1} + 4}} + \dfrac{1}{{\sqrt {6 – x} + 1}} + 3x + 1 > 0\\
\Rightarrow x = 5 \Rightarrow S = \left\{ 5 \right\}
\end{array}$