Question

[✉lãɴH⃣╰❥häńZ͜͡╰❥k̫ınżζ●]

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

a) Rút gọn A 

b) Tìm x biết A\(\ge\) \(\dfrac{1}{2}\)

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\\\sqrt{1+x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=2\) ta được:

\(A=\dfrac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\dfrac{\sqrt{\dfrac{a^2+b^2}{2}-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{a^2+b^2-ab}\)

\(=\sqrt{\dfrac{a^2+b^2-2ab}{2}}\left(a+b\right)=\dfrac{\left|a-b\right|\left(a+b\right)}{\sqrt{2}}\)

\(=\dfrac{\left|\sqrt{1-x}-\sqrt{1+x}\right|\left(\sqrt{1-x}+\sqrt{1+x}\right)}{\sqrt{2}}\)

- Với \(-1\le x\le0\Rightarrow A=\dfrac{\left(\sqrt{1-x}-\sqrt{1+x}\right)\left(\sqrt{1-x}+\sqrt{1+x}\right)}{\sqrt{2}}=-\sqrt{2}x\)

- Với \(0\le x\le1\Rightarrow A=\dfrac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\sqrt{2}}=\sqrt{2}x\)

b.

TH1: \(\left\{{}\begin{matrix}-1\le x\le0\\-\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow-1\le x\le-\dfrac{1}{2\sqrt{2}}\)

TH2: \(\left\{{}\begin{matrix}0\le x\le1\\\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2\sqrt{x}}\le x\le1\)