Toán Lớp 10: tìm m để pt $\sqrt{x^{2} -x+m}$ =$\sqrt{x-3}$ có nghiệm duy nhất
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Toán Lớp 10: tìm m để pt $\sqrt{x^{2} -x+m}$ =$\sqrt{x-3}$ có nghiệm duy nhất
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Dkxd:x \ge 3\\
\sqrt {{x^2} – x + m} = \sqrt {x – 3} \\
\Leftrightarrow {x^2} – x + m = x – 3\\
\Leftrightarrow {x^2} – 2x + m + 3 = 0\left( * \right)
\end{array}$
\Delta ‘ = 1 – m – 3 = – m – 2 = 0\\
\Leftrightarrow m = – 2\\
Khi:m = – 2\\
\Leftrightarrow {x^2} – 2x + 1 = 0\\
\Leftrightarrow {\left( {x – 1} \right)^2} = 0\\
\Leftrightarrow x = 1 < 3\left( {ktm} \right)
\end{array}$
{x_1} < 3 \le {x_2}\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} – 3 < 0\\
{x_2} – 3 \ge 0
\end{array} \right.\\
\Leftrightarrow \left( {{x_1} – 3} \right)\left( {{x_2} – 3} \right) \le 0\left( * \right)\\
Khi:\Delta ‘ > 0\\
\Leftrightarrow – m – 2 > 0\\
\Leftrightarrow m < – 2\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m + 3
\end{array} \right.\\
Do:\left( {{x_1} – 3} \right)\left( {{x_2} – 3} \right) \le 0\\
\Leftrightarrow {x_1}{x_2} – 3\left( {{x_1} + {x_2}} \right) + 9 \le 0\\
\Leftrightarrow m + 3 – 3.2 + 9 \le 0\\
\Leftrightarrow m + 6 \le 0\\
\Leftrightarrow m \le – 6\left( {tm} \right)\\
Vậy\,m \le – 6
\end{array}$