\(câu:10.4\)
\(x^2-2mx+m^2+m-6=0\left(1\right)có\) \(nghiệm\) \(dương\) \(duy\) \(nhất\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\Delta'=0\\x=-\dfrac{b}{2a}\in\left(0;+\text{∞}\right)\end{matrix}\right.\\x1\le0< x2\left(4\right)\end{matrix}\right.\left(3\right)\)
\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}\Delta'=m^2-m^2-m+6=0\Leftrightarrow m=6\left(5\right)\\x=\dfrac{2m}{2}=m=6\in\left(0;+\text{∞}\right)\left(thỏa\right)\end{matrix}\right.\)
\(\left(4\right)\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\\x1.x2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-m+6>0\\m^2+m-6\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 6\\-3\le m\le2\end{matrix}\right.\Leftrightarrow-3\le m\le2\left(6\right)\)
\(\left(5\right)\left(6\right)\Rightarrow m\in\left\{-3;-2;-1;0;1;2;6\right\}\Rightarrow D\) \(\)
\(câu\)\(11.1A\)
