Question

1．設(shè)正實(shí)數(shù)x，y滿足xy=$\frac{x+2y}{2x-4y}$，則實(shí)數(shù)x的最小值為$1+\sqrt{2}$．

Answer 1

分析 正實(shí)數(shù)x，y滿足xy=$\frac{x+2y}{2x-4y}$，變形為：4xy²+（2-2x²）y+x=0，可得：$\left\{\begin{array}{l}{△=（2-2{x}^{2}）^{2}-16{x}^{2}≥0}\\{\frac{2{x}^{2}-2}{4x}＞0}\\{\frac{x}{4x}＞0}\end{array}\right.$，解得即可得出．

解答 解：正實(shí)數(shù)x，y滿足xy=$\frac{x+2y}{2x-4y}$，
變形為：4xy²+（2-2x²）y+x=0，
∴$\left\{\begin{array}{l}{△=（2-2{x}^{2}）^{2}-16{x}^{2}≥0}\\{\frac{2{x}^{2}-2}{4x}＞0}\\{\frac{x}{4x}＞0}\end{array}\right.$，解得：x²≥3+2$\sqrt{2}$，
∴x$≥1+\sqrt{2}$．
則實(shí)數(shù)x的最小值為1+$\sqrt{2}$．
故答案為：$1+\sqrt{2}$．

點(diǎn)評(píng) 本題考查了一元二次不等式的解法，考查了推理能力與計(jì)算能力，屬于中檔題．