Question

Tìm GTNN,GTLN của biểu thức A=\(\frac{1}{2-\sqrt{3-x^2}}\)

Answer 1

Lời giải:
ĐK: $-\sqrt{3}\leq x\leq \sqrt{3}$

Ta thấy:

$x^2\geq 0, \forall x\in [-\sqrt{3}; \sqrt{3}]$

$\Rightarrow 3-x^2\leq 3\Rightarrow \sqrt{3-x^2}\leq \sqrt{3}$

$\Rightarrow 2-\sqrt{3-x^2}\geq 2-\sqrt{3}$

$\Rightarrow A=\frac{1}{2-\sqrt{3-x^2}}\leq \frac{1}{2-\sqrt{3}}=2+\sqrt{3}$

Vậy $A_{\max}=2+\sqrt{3}$ khi $x^2=0\Leftrightarrow x=0$

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$\sqrt{3-x^2}\geq 0$ với mọi $x\in [-\sqrt{3};\sqrt{3}]$

$\Rightarrow 2-\sqrt{3-x^2}\leq 2$

$\Rightarrow A=\frac{1}{2-\sqrt{3-x^2}}\geq \frac{1}{2}$

Vậy $A_{\min}=\frac{1}{2}$ khi $3-x^2=0\Leftrightarrow x=\pm \sqrt{3}$