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

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

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

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

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

$\Rightarrow A - 4 \geq 0$

$\Rightarrow A \geq 4$

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

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

\(P\le\sqrt{2\left(3x-5+7-3x\right)}=2\)

\(P_{max}=2\) khi \(3x-5=7-3x\Rightarrow x=2\)

\(A=2\left(x-1\right)+\dfrac{9}{x-1}+2\ge2\sqrt{\dfrac{18\left(x-1\right)}{x-1}}+2=6\sqrt{2}+2\)

\(A_{min}=6\sqrt{2}+2\) khi \(x=\dfrac{2+3\sqrt{2}}{2}\)

a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

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