■★均值不等式,MAX,MIN,最大值和最小值的使用法问题!☆●书中工式有 √(a^2+b^2)/2 >= (a+b)/2 >= √ab当求(a+b)/2的最小值时,可用a=b,则为(a+b)/2 min= √ab = a或b因为有a=b,所以同时有 (a+b)/2 max=√(a^2+
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/25 22:52:02
![■★均值不等式,MAX,MIN,最大值和最小值的使用法问题!☆●书中工式有 √(a^2+b^2)/2 >= (a+b)/2 >= √ab当求(a+b)/2的最小值时,可用a=b,则为(a+b)/2 min= √ab = a或b因为有a=b,所以同时有 (a+b)/2 max=√(a^2+](/uploads/image/z/4311504-0-4.jpg?t=%E2%96%A0%E2%98%85%E5%9D%87%E5%80%BC%E4%B8%8D%E7%AD%89%E5%BC%8F%2CMAX%2CMIN%2C%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%E7%9A%84%E4%BD%BF%E7%94%A8%E6%B3%95%E9%97%AE%E9%A2%98%21%E2%98%86%E2%97%8F%E4%B9%A6%E4%B8%AD%E5%B7%A5%E5%BC%8F%E6%9C%89+%E2%88%9A%28a%5E2%2Bb%5E2%29%2F2+%3E%3D+%28a%2Bb%29%2F2+%3E%3D+%E2%88%9Aab%E5%BD%93%E6%B1%82%EF%BC%88a%2Bb%EF%BC%89%2F2%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%97%B6%2C%E5%8F%AF%E7%94%A8a%3Db%2C%E5%88%99%E4%B8%BA%28a%2Bb%29%2F2+min%3D+%E2%88%9Aab+%3D+a%E6%88%96b%E5%9B%A0%E4%B8%BA%E6%9C%89a%3Db%2C%E6%89%80%E4%BB%A5%E5%90%8C%E6%97%B6%E6%9C%89+%28a%2Bb%29%2F2+max%3D%E2%88%9A%28a%5E2%2B)
xMPJAčL\&Fvg8.;0PK80P&D(C {n*!q禹r9pZٲͪUR'魧hr"%C_U\;G{tԯȠi=\7j;مQ%ȉBKG !ædX*S !%FK
"L?*R:DҠXHEJyx*IP<%ҁMcY?(=iC KW3"0i]$U
+e?B/^ӝlS&
X(P6CXѶr
gN9!ӅU )G@@a{п!vv3ɥ~Q