Question

12．已知x，y∈（-$\sqrt{2}$，$\sqrt{2}$），且xy=1，那么$\frac{2}{2-{x}^{2}}$+$\frac{4}{4-{y}^{2}}$的最小值是$\frac{16+4\sqrt{2}}{7}$．

Answer 1

分析 x、y∈（-$\sqrt{2}$，$\sqrt{2}$）且xy=1，可得y=$\frac{1}{x}$（x≠0）．化簡(jiǎn)變形$\frac{2}{2-{x}^{2}}$+$\frac{4}{4-{y}^{2}}$，利用基本不等式的性質(zhì)即可得出．

解答 解：∵x、y∈（-$\sqrt{2}$，$\sqrt{2}$）且xy=1，∴y=$\frac{1}{x}$（x≠0）．
則$\frac{2}{2-{x}^{2}}$+$\frac{4}{4-{y}^{2}}$=$\frac{2}{2-{x}^{2}}$+$\frac{4}{4-\frac{1}{{x}^{2}}}$
=$\frac{2}{2-{x}^{2}}$+$\frac{1}{{4x}^{2}-1}$+1
=1+$\frac{{7x}^{2}}{{9x}^{2}-{4x}^{4}-2}$
=1+$\frac{7}{9-（{4x}^{2}+\frac{2}{{x}^{2}}）}$≥1+$\frac{7}{9-2•2\sqrt{{2x}^{2}•\frac{1}{{x}^{2}}}}$
=$\frac{16+4\sqrt{2}}{7}$，當(dāng)且僅當(dāng)x²=$\frac{\sqrt{2}}{2}$，y²=$\sqrt{2}$時(shí)取等號(hào)．
∴$\frac{2}{2-{x}^{2}}$+$\frac{4}{4-{y}^{2}}$的最小值為$\frac{16+4\sqrt{2}}{7}$，
故答案為：$\frac{16+4\sqrt{2}}{7}$．

點(diǎn)評(píng) 本題考查了基本不等式的性質(zhì)，考查了變形能力、推理能力與計(jì)算能力，屬于中檔題．