Bài 2:
a) \(C=\dfrac{4-x^2}{x^2+1}=\dfrac{-\left(x^2+1\right)+5}{x^2+1}=-1+\dfrac{5}{x^2+1}\)
\(x^2+1\ge1\Rightarrow0< \dfrac{5}{x^2+1}\le5\)
\(\Rightarrow C\le-1+5=4\)
\(MaxC=4\Leftrightarrow x=0\)
b) \(D=\dfrac{x^2-4x-4}{x^2-4x+5}=\dfrac{\left(x^2-4x+5\right)-9}{x^2-4x+5}=1-\dfrac{9}{x^2-4x+5}\)
\(x^2-4x+5=\left(x-2\right)^2+1\ge1\)
\(\Rightarrow0< \dfrac{9}{x^2-4x+5}\le9\)
\(\Rightarrow0>-\dfrac{9}{x^2-4x+5}\ge-9\)
\(\Rightarrow D\ge1-9=-8\)
\(MinD=-8\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
Bài 3:
\(T=\dfrac{8x+12}{x^2+4}=\dfrac{\left(x^2+8x+16\right)-\left(x^2+4\right)}{x^2+4}=\dfrac{\left(x+4\right)^2}{x^2+4}-1\)
\(\left\{{}\begin{matrix}\left(x+4\right)^2\ge0\\x^2+4>0\end{matrix}\right.\Rightarrow\dfrac{\left(x+4\right)^2}{x^2+4}\ge0\)
\(\Rightarrow T\ge0-1=-1\)
\(MinT=-1\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)
\(T=\dfrac{8x+12}{x^2+4}=\dfrac{-\left(4x^2-8x+4\right)+4\left(x^2+4\right)}{x^2+4}=\dfrac{-4\left(x-1\right)^2}{x^2+4}+4\)
\(\left\{{}\begin{matrix}-4\left(x-1\right)^2\le0\\x^2+4>0\end{matrix}\right.\Rightarrow\dfrac{-4\left(x-1\right)^2}{x^2+4}\le0\)
\(\Rightarrow T\le0+4=4\)
\(MaxT=4\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
Bài 4:
\(A=\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\). Theo Cauchy cho 2 số dương:
\(\left\{{}\begin{matrix}x+y\ge2\sqrt{xy}>0\\\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}>0\end{matrix}\right.\)
Nhân vế theo vế của 2 BĐT trên ta được:
\(A=\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge4\)
\(MinA=4\Leftrightarrow x=y\)
