大家帮帮忙 急急急 十万火急！！！！ 100分！！！ 急急急！！！！！！！！十万火急！！

\u6025!!\u6025!!\u6025!!\u5341\u4e07\u706b\u6025!!!!!!!\u5341\u4e07\u706b\u6025!\u5927\u5bb6\u6765\u5e2e\u5e2e\u5fd9!!!!!

\u2460 2\u00d73\u00d74\u5206\u4e4b1\u52a03\u00d74\u00d75\u5206\u4e4b1\u52a04\u00d75\u00d76\u52a05\u00d76\u00d77\u5206\u4e4b1\u52a0\u202698\u00d799\u00d7100\u5206\u4e4b1
=\uff081/2*3-1/3*4)/2+(1/3*4-1/4*5)/2+(1/4*5-1/5*6)/2+(1/5*6-1/6*7)/2+...+(1/98*99-1/99*100)/2
=(1/2*3-1/3*4+1/3*4-1/4*5+1/4*5-1/5*6+1/5*6-1/6*7+...+1/98*99-1/99*100)/2
=(1/2*3-1/99*100)/2
=(1/6-1/9900)/2
=1649/19800


\u24611\u00d72\u00d73\u00d74\u5206\u4e4b1\u52a02\u00d73\u00d74\u00d75\u5206\u4e4b1\u52a0\u2026\u52a07\u00d78\u00d79\u00d710\u5206\u4e4b1
=(1/1*2*3-1/2*3*4)/3+(1/2*3*4-1/3*4*5)/3+...+(1/7*8*9-1/8*9*10)/3
=(1/1*2*3-1/8*9*10)/3
=(1/6-1/720)/3
=119/2160

Fan meizhong was called "Fan the fleer" because of his manners in Wenchuan earth quake.Recently,Fan became a teacher again.It raised a hot discussion in society.Personally,I am against of it.Fan's being a teacher just can meet the need of teaching but educating,because the true education is to picture a student who cannot only do well in the examnation but also have the determination and ability to help the others.In my heart ,a good teacher is the one who can develop his students overall.He has to pump the knowledge into the students brain in a good way,but also to instill the spirit of being offer to help others to theri students.

可以分别画出f(x)=sqrt(1+x^2),f(x)=sqrt(1+x),f(x)=sqrt(1-x)的函数曲线，其中第一个为双曲线的上半边，后两个为横着的半个抛物线。
从函数曲线，很容易的出，m>=0时，原函数最大值为无穷大。
m<0时，有最大值。此时，在x＝0点取最大值
g(m)=m+2(m<0)

可以化简，楼主你可以把式子化成f(x)=m*sqrt(1+x^2)+2+2x^2，令t=x^2就可以把式子大大简化，这时就有g(t)=f(x)=m*sqrt(1+t)+2t+2，求g(t)在t>=0时的最大值就行了，现在是不是很明了啊？

f(x)定义域为[-1,1]，且为偶函数，
要求最大值，只需考虑x>=0时的情况，因为f(-x)=f(x)，以下假设0<=x<=1
f'(x)=mx/√(1+x²)+1/(2√(1+x))-1/(2√(1-x))
当m<=0时，f'(x)<=1/(2√(1+x))-1/(2√(1-x))<=0，所以f(x)在[0,1]上为减函数，所以g(m)=f(0)=m+1+1=m+2
当m>0时，令f'(x)=0得
x^4+(2m+1)x^2-2m=0
令h(x)=x^4+(2m+1)x^2-2m，0<=x<=1，m>0，
h(0)=-2m<0，h(1)=2>0，显然h(x)=0只有一个小于1的正根，设这个正根为t，
t=√((√(4m²+12m+1)-2m-1)/2)
容易观察出当0<x<t时，f(x)为增函数，当t<x<1时，f(x)为减函数，
由f(x)的连续性可知，当m>0时，g(m)=f(t)=...太难化简了，你自己算吧。

当m<=0时，g(m)=m+2
当m>0时，g(m)=f(t), t=√((√(4m²+12m+1)-2m-1)/2)

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