（1）在区间 上用二分法求方程e^2+10X-2=0的近似根，要求误差不超过0.5*10^(-3) 。

老兄有没有搞错;方程是不是x^2+10x-2=0;

#include <math.h>
#include "iostream.h"
double fx(double x)
{
return exp(x)+10*x-2;
}
void main()
{
int count=0;
double a=0;
double b=1;
double xk=0;
cout<<"对分法："<<endl;
while (1)
{
xk=(a+b)/2;
if ((b-a)<=0.5*pow(10,-3))
{
cout<<"\n计算结果是:Xk="<<xk<<endl;
break;
}
if (fx(a)*fx(xk)<0)
{
b=xk;
}
else if (fx(a)*fx(xk)>0)
{
a=xk;
}
count++;
cout<<"\n第"<<count<<"次对分";
cout<<"\na的当前值"<<a;
cout<<"\nb的当前值:"<<b<<endl;
}
cout<<"对分次数:"<<count;
double stop;//暂停
//cin>>stop;
cout<<"牛顿迭代法："<<endl;//X(k+1)=X(k)-f(X(k))/(e(x)+