a, \(A=\dfrac{6}{x^2-2x+3}\)\(=\dfrac{6}{x^2-2x+1+2}=\dfrac{6}{\left(x-1\right)^2+2}\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\Leftrightarrow\left(x-1\right)^2+2\ge2\)
\(\Leftrightarrow\dfrac{1}{\left(x-1\right)^2+2}\le\dfrac{1}{2}\Leftrightarrow\dfrac{6}{\left(x-1\right)^2+2}\le3\)
Dấu bằng xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy MaxA = 3 khi x = 1
b, \(B=\dfrac{4}{x^2+6x+11}=\dfrac{4}{x^2+6x+9+2}=\dfrac{4}{\left(x+3\right)^2+2}\)
Ta có: \(\left(x+3\right)^2\ge0\forall x\Leftrightarrow\left(x+3\right)^2+2\ge2\)\(\Leftrightarrow\dfrac{1}{\left(x+3\right)^2+2}\le\dfrac{1}{2}\Leftrightarrow\dfrac{4}{\left(x+3\right)^2+2}\le2\)
Dấu bằng xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy MaxB = 2 khi x = -3
Bài 2:
\(A=\dfrac{5}{2x-x^2}=\dfrac{5}{-\left(x^2-2x+1\right)+1}=\dfrac{5}{-\left(x-1\right)^2+1}\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\Leftrightarrow-\left(x-1\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x-1\right)^2+1\le1\Leftrightarrow\dfrac{1}{-\left(x-1\right)^2+1}\ge1\)\(\Leftrightarrow\dfrac{5}{-\left(x-1\right)^2+1}\ge5\)
Dấu bằng xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy MinA = 5 khi x = 1
