Question

Tìm GTLN và GTNN của

a) A= \(\frac{3}{1+2\sqrt{3-x^2}}\)

b) B = \(\sqrt{9+4x-x^2}\)

Answer 1

a/ ĐKXĐ: ...

Ta có \(\sqrt{3-x^2}\ge0\Rightarrow A\le\frac{3}{1+2.0}=3\)

\(\Rightarrow A_{max}=3\) khi \(x=\pm\sqrt{3}\)

\(3-x^2\le3\Rightarrow\sqrt{3-x^2}\le\sqrt{3}\Rightarrow A\ge\frac{3}{1+2\sqrt{3}}\)

\(\Rightarrow A_{min}=\frac{3}{1+2\sqrt{3}}\) khi \(x=0\)

b/ĐKXĐ:...

\(B=\sqrt{13-\left(x^2-4x+4\right)}=\sqrt{13-\left(x-2\right)^2}\)

\(13-\left(x-2\right)^2\le13\Rightarrow B\le\sqrt{13}\)

\(\Rightarrow B_{max}=\sqrt{13}\) khi \(x=2\)

\(\sqrt{13-\left(x-2\right)^2}\ge0\Rightarrow B_{min}=0\) khi \(\left(x-2\right)^2=13\Rightarrow x=2\pm\sqrt{13}\)