?$解:甲兩次買菜的均價為\frac{3+2}{1+1}=2.5(元/千克)$?
?$乙兩次買菜的均價為\frac{3+3}{1+1.5}=2.4(元/千克)$?
?$[數(shù)學(xué)思考]\overline{x}_{甲}=\frac{am+bm}{m+m}=\frac{a+b}{2}(元/千克)$?
?$\overline{x}_{乙}=\frac{n+n}{\frac {n}{a}+\frac {n}}=\frac{2ab}{a+b}(元/千克)$?
?$\overline{x}_{甲}≥\overline{x}_{乙},理由:$?
?$\ \overline{x}_{甲}-\overline{x}_{乙}=\frac{(a+b)^{2}-4ab}{2(a+b)}=\frac{(a-b)^{2}}{2(a+b)}$?
?$∵a＞0,b＞0,(a-b)^{2}≥0,∴\frac{(a-b)^{2}}{2(a+b)}≥0$?
?$∴\overline{x}_{甲}≥\overline{x}_{乙}$?
?$[知識遷移]t_{1}＜t_{2}，理由:$?
?$∵t_{1}=\frac{s}{v}+\frac {s}{v}=\frac{2s}{v}$?
?$t_{2}=\frac {s}{v+p}+\frac {s}{v-p}=\frac {2sv}{v^{2}-p^{2}}$?
?$\ ∴t_{1}-t_{2}=\frac {2s(v^{2}-p^{2})-2sv^{2}}{v(v^{2}-p^{2})}=\frac {-2sp^{2}}{v(v^{2}-p^{2})}\ $?
?$∵s＞0,p＞0,v＞0,v＞p$?
?$∴ t_{1}-t_{2}＜0$?
?$∴t_{1}＜t_{2}$?
