\(M=\dfrac{4x+1}{x^2+3}\)
\(M+1=\dfrac{4x+1}{x^2+3}+\dfrac{x^2+3}{x^2+3}\)
\(M+1=\dfrac{x^2+4x+4}{x^2+3}=\dfrac{\left(x+2\right)^2}{x^2+3}\ge0\)
\(\Rightarrow M\ge-1\Leftrightarrow x=-2\)
Vậy MINM=-1<=>x=-2
C2:\(M=\dfrac{4x+1}{x^2+3}\)
\(\Leftrightarrow Mx^2+3M=4x+1\)
\(\Leftrightarrow Mx^2-4x+3M-1=0\left(1\right)\)
+)Xét M=0=>\(x=\dfrac{-1}{4}\)
+Xét \(M\ne0\)
=>Để pt(1) có nghiệm thì \(\Delta'=\left(-2\right)^2-M\left(3M-1\right)\ge0\)
\(\Leftrightarrow4-3M^2+M\ge0\)
\(\Leftrightarrow-1\le M\le\dfrac{4}{3}\)
\(\Rightarrow MINM=-1\Leftrightarrow x=-2\)
\(MAXM=\dfrac{4}{3}\Leftrightarrow x=\dfrac{3}{2}\)
Ta có : \(M=\dfrac{4x+1}{x^2+3}\)
\(=\dfrac{-\left(x^2+3\right)+x^2+4x+4}{x^2+3}\)
\(=-1+\dfrac{\left(x+2\right)^2}{x^2+3}\)
Vì \(\dfrac{\left(x+2\right)^2}{x^2+3}\ge0\) \(\Rightarrow M\ge-1\)
Dấu ''='' xảy ra \(\Leftrightarrow\) x = - 2
Vậy Min M = -1 \(\Leftrightarrow x=-2\)
