Question

A=\(\frac{\sqrt{1-\sqrt{1-x^2}}.\left[\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}\right]}{2-\sqrt{1-x^2}}\)

Rút gọn A

Answer 1

Lời giải:
ĐKXĐ:...........

Đặt $\sqrt{1+x}=a; \sqrt{1-x}=b(a\geq 0)\Rightarrow a^2+b^2=2$

\(A=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}.(a^3+b^3)}{a^2+b^2-ab}\)

\(=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}=\sqrt{\frac{(a-b)^2}{2}}.(a+b)=\frac{a-b}{\sqrt{2}}.(a+b)\)

\(=\frac{a^2-b^2}{\sqrt{2}}=\frac{1+x-(1-x)}{\sqrt{2}}=\frac{2x}{\sqrt{2}}=\sqrt{2}x\)

Answer 2

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