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\(y'=x^2-2mx+m\)
\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'\le0\end{matrix}\right.\Leftrightarrow m^2-m\le0\Leftrightarrow0\le m\le1\)
\(y'=x^2-2x+m\)
\(y'\ge0\) ; \(\forall x\in\left(1;3\right)\Leftrightarrow x^2-2x+m\ge0\) ;\(\forall x\in\left(1;3\right)\)
\(\Leftrightarrow m\ge\max\limits_{\left(1;3\right)}\left(-x^2+2x\right)\)
Xét hàm \(f\left(x\right)=-x^2+2x\) trên \(\left(1;3\right)\)
\(-\dfrac{b}{2a}=1\) ; \(f\left(1\right)=1\) ; \(f\left(3\right)=-3\)
\(\Rightarrow m\ge1\)
\(f'\left(x\right)=\dfrac{\left(2x-3\right)\left(x-1\right)-x^2+3x-7}{\left(x-1\right)^2}=\dfrac{x^2-2x-4}{\left(x-1\right)^2}\)
\(f'\left(x\right)=0\Leftrightarrow x^2-2x-4=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow x_1^2+x_2^2=12\)
Hoặc bạn dùng Vi-ét cũng được, tùy
\(y'=\dfrac{-m^2-1}{\left(x-m\right)^2}\)
\(y'< 0\) ;\(\forall x\in\left(0;1\right)\Leftrightarrow\left[{}\begin{matrix}m\ge1\\m\le0\end{matrix}\right.\)
\(y'=\dfrac{-2m-1}{\left(x-2\right)^2}\)
\(y'< 0\) với mọi x thuộc TXĐ \(\Leftrightarrow-2m-1< 0\Leftrightarrow m>-\dfrac{1}{2}\)
\(y'=-3x^2-6x+m\Rightarrow y''=-6x-6\)
\(y''=0\Leftrightarrow-6x-6=0\Leftrightarrow x=-1\notin\left[0;1\right]\)
\(\left\{{}\begin{matrix}y'\left(0\right)=m\\y'\left(1\right)=m-9\end{matrix}\right.\Rightarrow^{max}_{\left[0;1\right]}y'=y'\left(0\right)=m\)
\(\Rightarrow m=10\)
\(y'=3x^2-2x+m\)
\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'\le0\end{matrix}\right.\Leftrightarrow1-3m\le0\Leftrightarrow m\ge\dfrac{1}{3}\)
\(y'=4mx^3+2mx=2mx\left(2x^2+1\right)\)
Do \(2x\left(x^2+1\right)>0\) ;\(\forall x>0\)
\(\Rightarrow y'\ge0\) ;\(\forall x>0\) khi và chỉ khi \(m>0\)
\(y'=\dfrac{\left(2x-m\right)\left(x^2+1\right)-2x\left(x^2-mx+m\right)}{\left(x^2+1\right)^2}=\dfrac{2x-mx^2-m+2mx^2-2mx}{\left(x^2+1\right)^2}=\dfrac{mx^2+2\left(1-m\right)x-m}{\left(x^2+1\right)^2}\)
\(y'=0\Leftrightarrow mx^2+2\left(1-m\right)x-m=0\)
Xet \(m=0\) ko thoa man pt
Xet \(m\ne0\)
\(\left\{{}\begin{matrix}\Delta'>0\\\dfrac{2\left(m-1\right)}{m}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(1-m\right)^2+m^2>0\left(ld\right)\\m=-2\end{matrix}\right.\Rightarrow m=-2\)