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void main(){

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int x;
float fx;
start:
printf("請(qǐng)輸入x值(x值大于或等于0):");
scanf("%d",x);
if(x0)
goto start;
fx=x*x+1.0/x+sqrt(x);
printf(fx=%.3lf\n",fx);
}
#include?stdio.h
#include?math.h
double?f(double?x);
int?main(void)?
{
double?x;
scanf("%lf",x);
printf("%lf\n",f(x));
return?0;
}
double?f(double?x)
{
return?sqrt((x-1)*(x-2))/x;
}
#include?stdio.h
float?fun(float?x)
{
if(x0)?return?1;
else?if(x==0)?return?0;
else?return?-1;
}
int?main()
{
float?x;
scanf("%f",x);
printf("%f",fun(x));
return?0;
}
#include "stdio.h"
#include "math.h"
/*求一元二次方程ax*x+bx+c=0的解*/
main()
{
float a,b,c,d,e,f,k,n,m;
printf("請(qǐng)輸入X1的值:");
scanf("%f",a);
printf("請(qǐng)輸入Y1的值:");
scanf("%f",b);
printf("請(qǐng)輸入X2的值:");
scanf("%f",c);
printf("請(qǐng)輸入Y2的值:");
scanf("%f",d);
printf("請(qǐng)輸入X3的值:");
scanf("%f",e);
printf("請(qǐng)輸入Y3的值:");
scanf("%f",f);
k=(a*c*f*(c-a)+b*c*e*(e-c)+a*d*e*(a+e))/(e*e*(c-a)+a*a*(e-c)+c*c*(a-e));
n=(a*a*(k-d) -c*c*(k-d))/(c*c*a-a*a*c);
m=(b-n*a-k)/(a*a);
{
printf("k=%f\n",k);
printf("n=%f\n",n);
printf("m=%f\n",m);
printf("該二次函數(shù)解析式為:y=%fx2+%fx+%f\n",k,n,m);
}
}