Question

Tìm GTNN $A=\dfrac{5-3x}{\sqrt{1-x^2}}$

Answer 1

$A =\dfrac{5 - 3x}{\sqrt{1 - x^2}}\qquad (-1 < x < 1)$

$\Rightarrow A - 4 =\dfrac{5 -3x}{\sqrt{1 - x^2}}- 4$

$\Rightarrow A - 4 =\dfrac{5 - 3x - 4\sqrt{1-x^2}}{\sqrt{1 - x^2}}$

$\Rightarrow A - 4 =\dfrac{4(1-x) - 2.2\sqrt{1-x}.\sqrt{1+x} + 1 +x}{\sqrt{1 - x^2}}$

$\Rightarrow A - 4 =\dfrac{(2\sqrt{1-x} - \sqrt{1+x})^2}{\sqrt{1-x^2}}\geq 0$

$\Rightarrow A - 4 \geq 0$

$\Rightarrow A \geq 4$

Dấu $=$ xảy ra $\Leftrightarrow 2\sqrt{1-x}=\sqrt{1+x}\Leftrightarrow x =\dfrac35$

Vậy $\min A = 4\Leftrightarrow x =\dfrac35$