Question

13．已知a，b，c，d，x，y，z，w是互不相等的非零实数，且$\frac{{a}_{2}{b}^{2}}{{a}^{2}{y}^{2}+{b}^{2}{x}^{2}}$=$\frac{{b}^{2}{c}^{2}}{{b}^{2}{z}^{2}+{c}^{2}{y}^{3}}$=$\frac{{c}^{2}{d}^{2}}{{c}^{2}{w}^{2}+{d}^{2}{z}^{2}}$=$\frac{abcd}{xyzw}$，求$\frac{{a}^{2}}{{x}^{2}}+\frac{{b}^{2}}{{y}^{2}}+\frac{{c}^{2}}{{z}^{2}}+\frac{{d}^{2}}{{w}^{2}}$的值．

Answer 1

分析 设$\frac{{a}_{2}{b}^{2}}{{a}^{2}{y}^{2}+{b}^{2}{x}^{2}}$=$\frac{{b}^{2}{c}^{2}}{{b}^{2}{z}^{2}+{c}^{2}{y}^{3}}$=$\frac{{c}^{2}{d}^{2}}{{c}^{2}{w}^{2}+{d}^{2}{z}^{2}}$=$\frac{abcd}{xyzw}$=$\frac{1}{k}$，从而得$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{w}^{2}}{{d}^{2}}$+$\frac{{z}^{2}}{{c}^{2}}$=$\frac{xyzw}{abcd}$=k，即$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$、$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{w}^{2}}{{d}^{2}}$、$\frac{xyzw}{abcd}$=k，设$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$=k₁、$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{w}^{2}}{{d}^{2}}$=k₂，从而得k=$\sqrt{{{k}_{1}}^{2}•{{k}_{2}}^{2}}$、k₁+k₂=k，代入即可得．

解答 解：设$\frac{{a}_{2}{b}^{2}}{{a}^{2}{y}^{2}+{b}^{2}{x}^{2}}$=$\frac{{b}^{2}{c}^{2}}{{b}^{2}{z}^{2}+{c}^{2}{y}^{3}}$=$\frac{{c}^{2}{d}^{2}}{{c}^{2}{w}^{2}+{d}^{2}{z}^{2}}$=$\frac{abcd}{xyzw}$=$\frac{1}{k}$，
则$\frac{{a}^{2}{y}^{2}+{b}^{2}{x}^{2}}{{a}^{2}{b}^{2}}$=$\frac{{b}^{2}{z}^{2}+{c}^{2}{y}^{2}}{{b}^{2}{c}^{2}}$=$\frac{{c}^{2}{w}^{2}+{d}^{2}{z}^{2}}{{c}^{2}{d}^{2}}$=$\frac{xyzw}{abcd}$=k，
整理，得：$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{w}^{2}}{{d}^{2}}$+$\frac{{z}^{2}}{{c}^{2}}$=$\frac{xyzw}{abcd}$=k，
∴$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$、$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{w}^{2}}{{d}^{2}}$，$\frac{xyzw}{abcd}$=k，
设$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$=k₁，$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{w}^{2}}{{d}^{2}}$=k₂，
由$\frac{xyzw}{abcd}$=k可得k=$\sqrt{{{k}_{1}}^{2}•{{k}_{2}}^{2}}$，
由$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=$\frac{{a}^{2}{y}^{2}+{b}^{2}{x}^{2}}{{a}^{2}{b}^{2}}$得k₁+k₂=k，
∴原式=2×$\frac{1}{{k}_{1}}$+2×$\frac{1}{{k}_{2}}$=$\frac{2（{k}_{1}+{k}_{2}）}{{k}_{1}{k}_{2}}$=2．

点评 本题主要考查分式的化简求值，熟练掌握分式的运算法则和性质是解题的关键．