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重庆市2023-2024学年高二年级(下)2月月度质量检测数学.考卷答案试卷答案
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分析由1≤lg（xy）≤4，-1$≤lg\frac{x}{y}$≤2，可得：1≤lgx+lgy≤4，-1≤lgx-lgy≤2，而lg$\frac{{x}^{2}}{y}$=2lgx-lgy，设2lgx-lgy=m（lgx+lgy）+n（lgx-lgy），利用“待定系数法”即可得出．
解答解：由1≤lg（xy）≤4，-1$≤lg\frac{x}{y}$≤2，可得：1≤lgx+lgy≤4，-1≤lgx-lgy≤2，
而lg$\frac{{x}^{2}}{y}$=2lgx-lgy
设2lgx-lgy=m（lgx+lgy）+n（lgx-lgy），
∴$\left\{\begin{array}{l}{m+n=2}\\{m-n=-1}\end{array}\right.$，
解得m=$\frac{1}{2}$，n=$\frac{3}{2}$．
∴lg$\frac{{x}^{2}}{y}$=2lgx-lgy=$\frac{1}{2}$（lgx+lgy）+$\frac{3}{2}$（lgx-lgy），
∴-1≤$lg\frac{{x}^{2}}{y}$≤5，
故答案为：[-1，5]．
点评本题考查了不等式的性质、对数的运算法则，考查了推理能力与计算能力，属于中档题．
