Question

3．若實數(shù)x，y滿足3x+1≥e^x+y-3+e^2x-y+2則x+y=3．

Answer 1

分析 根據(jù)[（x+y-3）+1]≤e^x+y-3，[（2x-y+2）+1]≤e^2x-y+2，得到$\left\{\begin{array}{l}{x+y-2{≤e}^{x+y-3}}\\{2x-y+3{=e}^{2x-y+2}}\end{array}\right.$，求出x，y的值即可．

解答 解：令f（x）=e^x-（x+1），
則f′（x）=e^x-1，
令f′（x）＞0，解得：x＞0，
令f′（x）＜0，解得：x＜0，
故f（x）的最小值是f（0）=0，
故e^x≥x+1，
∵3x+1≥e^x+y-3+e^2x-y+2，
∴[（x+y-3）+2]+[（2x-y+2）+1]
≥e^x+y-3+e^2x-y+2，
即[（x+y-3）+1]≤e^x+y-3，[（2x-y+2）+1]≤e^2x-y+2，
∴$\left\{\begin{array}{l}{x+y-2{≤e}^{x+y-3}}\\{2x-y+3{=e}^{2x-y+2}}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x+y-3=0}\\{2x-y+2=0}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x=\frac{1}{3}}\\{y=\frac{8}{3}}\end{array}\right.$，
∴x+y=3，
故答案為：3．

點評 本題考查了不等式的性質，考查轉化思想，是一道綜合題．