解方程组2X+3Y=64X-Z=10Z=2解不等式组3x-4＜5×＋34×＋1＞2＋3×

Question

解方程组2X+3Y=6 4X-Z=10 Z=2 解不等式组3x-4＜5×＋3 4×＋1＞2＋3×

Answer 1

解为1
/*
牛顿迭代法解方程组的解
x0为迭代的初值,n为迭代次数,jingdu为精度
function为求根代数式,d2functoin为其导数
返回最终符合一定精度的根
*/
double newton_diedai(double x0,int *n,double jingdu)
{
double x,temp;
temp=d2function(x0);
if (fabs(temp)>1e-10) /*防止除数为0*/
{
x=x0-function(x0)/temp;
printf("n=%d\tx=%.5lf\n",*n,x);
}
else
{
printf("error:div 0:\nPress any key to exit:");
getch();
exit(1);
}
if (++(*n)>MAX_DIEDAI_TIME)
{
printf("diedai time:%d > MAX_DIEDAI_TIME:\nPress any key to exit:",*n);
getch();
exit(1);
}
temp=function(x);
if (fabs(temp)return x;
else
return newton_diedai(x,n,jingdu);
}