Rút gọn :
\(M=\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]}{2+\sqrt{1-x^2}}\)
@Phùng Khánh Linh
Lời giải:
Đặt \((\sqrt{1+x}=a; \sqrt{1-x}=b)\)
\(\Rightarrow a^2+b^2=2\) và \(a^2-b^2=2x\)
Khi đó:
\(M=\frac{\sqrt{1+ab}(a^3-b^3)}{2+ab}=\frac{\sqrt{1+ab}(a-b)(a^2+ab+b^2)}{a^2+b^2+ab}\)
\(=\sqrt{1+ab}(a-b)\)
\(=\sqrt{\frac{a^2+b^2}{2}+ab}(a-b)=\sqrt{\frac{a^2+b^2+2ab}{2}}(a-b)\)
\(=\sqrt{\frac{(a+b)^2}{2}}(a-b)=\frac{(a+b)(a-b)}{\sqrt{2}}=\frac{a^2-b^2}{\sqrt{2}}=\frac{2x}{\sqrt{2}}=\sqrt{2}x\)
\(M=\dfrac{\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]}{2+\sqrt{1-x^2}}\)
\(\Leftrightarrow M=\dfrac{\sqrt{2}.\sqrt{1+\sqrt{1-x^2}}\left[\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)
\(\Leftrightarrow M=\dfrac{\sqrt{2+2\sqrt{1-x^2}}\left[(\sqrt{\left(1+x\right)})^3-(\sqrt{\left(1-x\right)})^3\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)
\(\Leftrightarrow M=\dfrac{\sqrt{\left(1-x\right)+2\sqrt{\left(1-x\right)\left(1+x\right)}+(1+x)}.\left[(\sqrt{1+x})^3-\left(\sqrt{1-x}\right)^3\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)
\(\Leftrightarrow M=\dfrac{\sqrt{(\sqrt{1+x}+\sqrt{1-x})^2}.\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[\left(\sqrt{1+x}\right)^2+\sqrt{1+x}\sqrt{1-x}+\left(\sqrt{1-x}^2\right)\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)
\(\Leftrightarrow M=\dfrac{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[1+x+\sqrt{1-x^2}+1-x\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)
\(\Leftrightarrow M=\dfrac{(1+x-1+x)\left[2+\sqrt{1-x^2}\right]}{\sqrt{2}.(2+\sqrt{1-x^2})}\)
\(\Leftrightarrow M=\dfrac{2x}{\sqrt{2}}\)
\(\Leftrightarrow M=\sqrt{2}x\)
