Question

13．已知a，b，c，d，x，y，z，w是互不相等的非零實數(shù)，且$\frac{{a}_{2}^{2}}{{a}^{2}{y}^{2}+^{2}{x}^{2}}$=$\frac{^{2}{c}^{2}}{^{2}{z}^{2}+{c}^{2}{y}^{3}}$=$\frac{{c}^{2}p9vv5xb5^{2}}{{c}^{2}{w}^{2}+p9vv5xb5^{2}{z}^{2}}$=$\frac{abcd}{xyzw}$，求$\frac{{a}^{2}}{{x}^{2}}+\frac{^{2}}{{y}^{2}}+\frac{{c}^{2}}{{z}^{2}}+\frac{p9vv5xb5^{2}}{{w}^{2}}$的值．

Answer 1

分析 設$\frac{{a}_{2}^{2}}{{a}^{2}{y}^{2}+^{2}{x}^{2}}$=$\frac{^{2}{c}^{2}}{^{2}{z}^{2}+{c}^{2}{y}^{3}}$=$\frac{{c}^{2}p9vv5xb5^{2}}{{c}^{2}{w}^{2}+p9vv5xb5^{2}{z}^{2}}$=$\frac{abcd}{xyzw}$=$\frac{1}{k}$，從而得$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$+$\frac{{y}^{2}}{^{2}}$=$\frac{{w}^{2}}{p9vv5xb5^{2}}$+$\frac{{z}^{2}}{{c}^{2}}$=$\frac{xyzw}{abcd}$=k，即$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$、$\frac{{y}^{2}}{^{2}}$=$\frac{{w}^{2}}{p9vv5xb5^{2}}$、$\frac{xyzw}{abcd}$=k，設$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$=k₁、$\frac{{y}^{2}}{^{2}}$=$\frac{{w}^{2}}{p9vv5xb5^{2}}$=k₂，從而得k=$\sqrt{{{k}_{1}}^{2}•{{k}_{2}}^{2}}$、k₁+k₂=k，代入即可得．

解答 解：設$\frac{{a}_{2}^{2}}{{a}^{2}{y}^{2}+^{2}{x}^{2}}$=$\frac{^{2}{c}^{2}}{^{2}{z}^{2}+{c}^{2}{y}^{3}}$=$\frac{{c}^{2}p9vv5xb5^{2}}{{c}^{2}{w}^{2}+p9vv5xb5^{2}{z}^{2}}$=$\frac{abcd}{xyzw}$=$\frac{1}{k}$，
則$\frac{{a}^{2}{y}^{2}+^{2}{x}^{2}}{{a}^{2}^{2}}$=$\frac{^{2}{z}^{2}+{c}^{2}{y}^{2}}{^{2}{c}^{2}}$=$\frac{{c}^{2}{w}^{2}+p9vv5xb5^{2}{z}^{2}}{{c}^{2}p9vv5xb5^{2}}$=$\frac{xyzw}{abcd}$=k，
整理，得：$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$+$\frac{{y}^{2}}{^{2}}$=$\frac{{w}^{2}}{p9vv5xb5^{2}}$+$\frac{{z}^{2}}{{c}^{2}}$=$\frac{xyzw}{abcd}$=k，
∴$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$、$\frac{{y}^{2}}{^{2}}$=$\frac{{w}^{2}}{p9vv5xb5^{2}}$，$\frac{xyzw}{abcd}$=k，
設$\frac{{x}^{2}}{{a}^{2}}$=$\frac{{z}^{2}}{{c}^{2}}$=k₁，$\frac{{y}^{2}}{^{2}}$=$\frac{{w}^{2}}{p9vv5xb5^{2}}$=k₂，
由$\frac{xyzw}{abcd}$=k可得k=$\sqrt{{{k}_{1}}^{2}•{{k}_{2}}^{2}}$，
由$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{^{2}}$=$\frac{{a}^{2}{y}^{2}+^{2}{x}^{2}}{{a}^{2}^{2}}$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．

點評 本題主要考查分式的化簡求值，熟練掌握分式的運算法則和性質(zhì)是解題的關(guān)鍵．