Rút gọn: $\frac{\sqrt{1-\sqrt{1-x^{2}}}.(\sqrt{(1+x)^{3}}+\sqrt{(1-x)^3} )}{2-\sqrt{1-x^{2}}}$ Ai tl mk vote 5* + ctlhn nha

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1. 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:

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

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

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

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

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

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

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

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

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