Đáp án: $\dfrac{\sqrt{1-\sqrt{1-x^2}}\cdot (\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}=|x|\cdot \sqrt{2}$

Giải thích các bước giải:

Ta có:

$\sqrt{1-\sqrt{1-x^2}}=\sqrt{\dfrac12(2-2\sqrt{1-x^2})}$

$\to \sqrt{1-\sqrt{1-x^2}}=\sqrt{\dfrac12(2-2\sqrt{(1-x)(1+x)})}$

$\to \sqrt{1-\sqrt{1-x^2}}=\sqrt{\dfrac12((1+x)-2\sqrt{1+x}\cdot \sqrt{1-x}+(1-x))}$

$\to \sqrt{1-\sqrt{1-x^2}}=\sqrt{\dfrac12(\sqrt{1+x}- \sqrt{1-x})^2}$

$\to \sqrt{1-\sqrt{1-x^2}}=\dfrac{1}{\sqrt{2}}\cdot |\sqrt{1+x}-\sqrt{1-x}|$

Ta có:

$\sqrt{(1+x)^3}+\sqrt{(1-x)^3}=(\sqrt{1+x})^3+(\sqrt{1-x})^3$

$\to \sqrt{(1+x)^3}+\sqrt{(1-x)^3}=(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x})^2-\sqrt{1+x}\cdot \sqrt{1-x}+(\sqrt{1-x})^2)$

$\to \sqrt{(1+x)^3}+\sqrt{(1-x)^3}=(\sqrt{1+x}+\sqrt{1-x})(1+x-\sqrt{1-x^2}+1-x)$

$\to \sqrt{(1+x)^3}+\sqrt{(1-x)^3}=(\sqrt{1+x}+\sqrt{1-x})(2-\sqrt{1-x^2})$

Khi đó:

$A=\dfrac{\sqrt{1-\sqrt{1-x^2}}\cdot (\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}$

$\to A=\dfrac{\dfrac{1}{\sqrt{2}}\cdot |\sqrt{1+x}-\sqrt{1-x}|\cdot (\sqrt{1+x}+\sqrt{1-x})(2-\sqrt{1-x^2})}{2-\sqrt{1-x^2}}$

$\to A=\dfrac{1}{\sqrt{2}}\cdot |\sqrt{1+x}-\sqrt{1-x}|\cdot (\sqrt{1+x}+\sqrt{1-x})$

$\to A=\dfrac{1}{\sqrt{2}}\cdot |\sqrt{1+x}-\sqrt{1-x}|\cdot |\sqrt{1+x}+\sqrt{1-x}|$

$\to A=\dfrac{1}{\sqrt{2}}\cdot |(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})|$

$\to A=\dfrac{1}{\sqrt{2}}\cdot |1+x- (1-x)|$

$\to A=\dfrac{1}{\sqrt{2}}\cdot |2x|$

$\to A=|x|\cdot \sqrt{2}$