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Đặt \(2^x=t>0\Rightarrow t^2-mt+10-m=0\) (1)
Để pt đã cho có 2 nghiệm pb thì (1) có 2 nghiệm dương phân biệt
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=m^2-4\left(10-m\right)>0\\S=m>0\\P=10-m>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< -2-2\sqrt{11}\\m>-2+2\sqrt{11}\end{matrix}\right.\\0< m< 10\end{matrix}\right.\)
\(\Rightarrow m=\left\{5;6;7;8;9\right\}\) \(\Rightarrow\sum m=35\)
\(y'=x^2-\left(3m+2\right)x+2m^2+3m+1\)
\(\Delta=\left(3m+2\right)^2-4\left(2m^2+3m+1\right)=m^2\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{3m+2+m}{2}=2m+1\\x_2=\frac{3m+2-m}{2}=m+1\end{matrix}\right.\)
Để hàm số có cực đại, cực tiểu \(\Rightarrow x_1\ne x_2\Rightarrow m\ne0\)
- Nếu \(m>0\Rightarrow2m+1>m+1\Rightarrow\left\{{}\begin{matrix}x_{CĐ}=m+1\\x_{CT}=2m+1\end{matrix}\right.\)
\(\Rightarrow3\left(m+1\right)^2=4\left(2m+1\right)\) \(\Rightarrow3m^2-2m-1=0\Rightarrow\left[{}\begin{matrix}m=1\\m=-\frac{1}{3}< 0\left(l\right)\end{matrix}\right.\)
- Nếu \(m< 0\Rightarrow m+1>2m+1\Rightarrow\left\{{}\begin{matrix}x_{CĐ}=2m+1\\x_{CT}=m+1\end{matrix}\right.\)
\(\Rightarrow3\left(2m+1\right)^2=4\left(m+1\right)\Rightarrow12m^2+8m-1=0\)
\(\Rightarrow\left[{}\begin{matrix}m=\frac{-2+\sqrt{7}}{6}>0\left(l\right)\\m=\frac{-2-\sqrt{7}}{6}\end{matrix}\right.\) \(\Rightarrow\sum m=\frac{4-\sqrt{7}}{6}\)
\(\Leftrightarrow\max\limits_{\left[0;3\right]}f\left(x\right)\le16\)
Trên \(\left[0;3\right]\) xét hàm \(g\left(x\right)=x^3-3x+m\Rightarrow g'\left(x\right)=3x^2-3=0\Rightarrow x=1\)
Ta có: \(g\left(0\right)=m;\) \(g\left(1\right)=m-2\) ; \(g\left(3\right)=m+21\)
\(\Rightarrow\max\limits_{\left[0;3\right]}f\left(x\right)=max\left\{\left|m-2\right|;\left|m+18\right|\right\}\)
TH1: \(\left\{{}\begin{matrix}\left|m-2\right|\ge\left|m+18\right|\\\left|m-2\right|\le16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\le-8\\-14\le m\le18\end{matrix}\right.\)
\(\Rightarrow-14\le m\le-8\)
TH2: \(\left\{{}\begin{matrix}\left|m+18\right|\ge\left|m-2\right|\\\left|m+18\right|\le16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ge-8\\-34\le m\le-2\end{matrix}\right.\)
\(\Rightarrow-8\le m\le-2\)
Vậy \(-14\le m\le-2\)
