HS xác định \(\Leftrightarrow\left(m-10\right)x^2-2\left(m-10\right)x+1\ge0\)
Đặt \(f\left(x\right)=\left(m-10\right)x^2-2\left(m-10\right)x+1\)
\(f\left(x\right)\ge0,\forall x\in R\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m-10>0\\\left[-2\left(m-10\right)\right]^2-4\left(m-10\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>10\\4\left(m^2-20m+100\right)-4m+40\le0\end{matrix}\right.\)
\(\Leftrightarrow4m^2-80m+400-4m+40\le0\)
\(\Leftrightarrow4m^2-84m+440\le0\)
\(\Rightarrow\left\{{}\begin{matrix}m>10\\m\in\left[10;11\right]\end{matrix}\right.\)
\(KL:m\in(10;11]\)
- Với \(m=10\) thỏa mãn
- Với \(m\ne10\) hàm xác định trên R khi:
\(\left(m-10\right)x^2-2\left(m-10\right)+1\ge0;\forall x\)
\(\Leftrightarrow\left\{{}\begin{matrix}m-10>0\\\Delta'=\left(m-10\right)^2-\left(m-10\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>10\\\left(m-10\right)\left(m-11\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>10\\10\le m\le11\end{matrix}\right.\) \(\Rightarrow10< m\le11\)
Kết hợp lại \(\Rightarrow10\le m\le11\) (A)
