a) \(A=5-\sqrt{x^2-6x+14}\)
\(A=5-\sqrt{x^2-6x+9+5}\)
\(A=5-\sqrt{\left(x-3\right)^2+5}\le5-\sqrt{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Rightarrow x=3\)
Vậy maxA = \(5-\sqrt{5}\Leftrightarrow x=3\)
b) \(B=\frac{1}{\sqrt{x^2-4x+5}}\)
\(B=\frac{1}{\sqrt{x^2-4x+4+1}}\)
\(B=\frac{1}{\sqrt{\left(x-2\right)^2+1}}\le\frac{1}{1}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Rightarrow x=2\)
Vậy maxA = \(1\Leftrightarrow x=2\)
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