theo bai ra ta co\(y-4\le0;y+1\ge0\)
\(\Leftrightarrow\left(y-4\right)\left(y+1\right)\le0\)
\(\Leftrightarrow y^2-3y-4\le0\)
\(y=\dfrac{mx+n}{x^2+1}\Leftrightarrow x^2y-mx+y-n=0\)
pt co nghiem khi \(\Delta=m^2-4\left(y-n\right)y\)
\(=-\left(4y^2-4ny-m^2\right)\ge0\)
\(\Leftrightarrow y^2-ny-\dfrac{m^2}{4}\le0\)
dong nhat n=3;m=±4
\(y=\dfrac{mx+n}{x^2+1}\Rightarrow x^2y+y-mx-n=0\)
\(\Delta=m^2-4y\left(y-n\right)\ge0\)
\(\Leftrightarrow-4y^2+4yn+m^2\ge0\)
\(\Leftrightarrow4y^2-4yn+m^2\le0\)
Có: \(\left(y-4\right)\left(y+1\right)\le0\) với \(-1\le y\le4\)
\(\Rightarrow4\left(y-n\right)\left(y+\dfrac{m^2}{4}\right)\le0\)
\(\Rightarrow n=3,m=\pm2\)
bài này rất dể nhưng lại không ai thèm giải nên mk làm luôn nhé
bài làm :
bài này không cần điều kiện xác định
+) ta có : \(y_{max}=4\) \(\Rightarrow\left\{{}\begin{matrix}y\le4,\forall x\in R\\\exists x_0\in R,y\left(x_0\right)=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{mx+n}{x^2+1}\le4,\forall x\in R\\\exists x_0\in R,\dfrac{mx_0+n}{x_0^2+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-mx+\left(4-n\right)\ge0,\forall x\in R\\\exists x_0\in R,4x_0^2-mx_0+\left(4-n\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\Delta_{\left(1\right)}\le0\\\Delta_{\left(1\right)}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\Delta_{\left(1\right)}=0\Leftrightarrow m^2-16\left(4-n\right)=0\) ........(*)
tương tự ta có :
\(y_{min}=-1\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+mx+\left(n+1\right)\ge0,\forall x\in R\\\exists x_0\in R,x_0^2+mx_0+\left(n+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta_{\left(2\right)}\le0\\\Delta_{\left(2\right)}\ge0\end{matrix}\right.\) \(\Leftrightarrow\Delta_{\left(2\right)}=0\Leftrightarrow m^2-4\left(n+1\right)=0\) ........(**)
từ (*) và (**) ta có hệ : \(\left\{{}\begin{matrix}m^2-16\left(4-n\right)=0\\m^2-4\left(n+1\right)=0\end{matrix}\right.\)
đặc \(k=m^2\left(k\ge0\right)\) khi đó hệ \(\Leftrightarrow\left\{{}\begin{matrix}k+16n=64\\k-4n=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}k=16\\n=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2=16\\n=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m=4\\n=3\end{matrix}\right.\\\left\{{}\begin{matrix}m=-4\\n=3\end{matrix}\right.\end{matrix}\right.\)
vậy \(\left(m;n\right)\in\left\{\left(4;3\right);\left(-4;3\right)\right\}\)
