printf(“x1=%8.4f,x2=%8.4f\n”,x1,x2);给解释下什么意识!

printf(“x1=%8.4f,x2=%8.4f ”,x1,x2);给解释下什么意识! printf(该一元二次方程有两个解,x1 = %f,x2 = %f ,x1,x2); 在引号后面为什么还要写一次x1,x2? 杭电2001.为什么是wrong answer?#include#includemain(){float x1,x2,y1,y2;while(scanf(%f %f %f %f,&x1,&x2,&y1,&y2)!=EOF){printf(%.2f,sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));putchar(' ');}} printf(x1=%5.2f ,x1)中的5. f(x1.x2)=f(x1)+f(x2)证明奇偶性 证明f(x1+x2)+f(x1-x2)=2f(x1)f(x2)为偶函数 c语言编写一元2次方程.#include#includevoid main(){double a,b,c,x1,x2,d;scanf(%f,%f,%f,&a,&b,&c);d=b*b-4*a*c;x1=(-b+sqrt(d))/2;x2=(-b-sqrt(d))/2;if(d>0)printf(x1=%f,x2=%f ,x1,x2);if(d=0)printf(x1=x2=%f ,x1);if(d 用迭代法 求方程x=cosx得根,是误差小于10的负六.{double x1 x2;x1 = 0.0;x2 = cos(x1);while fabs(x2-x1) > (1e-6）{ x1 = x2; x2 = cos(x1);}printf (x=%f ,x2);}求解题思路 我看不懂啊 C语言编一元二次方程出了点问题,int main(int argc,char *argv[]){int a,b,c,x1,x2;printf(Please input:a,b,c );scanf(%f%f%f,&a,&b,&c);x1=b*b-4*a*c;if(x10){ x2=(-b+sqrt(x1))/(2*a);x1=(-b+sqrt(x1))/(2*a);printf(x1=%f x2=%f,x1,x2); } el 已知f(x)对任意实数x1 x2都有f(x1+x2)+f(x1-x2)=2f(x1)·f(x2) 求证f(x)为偶函数请各位看以下解法是否正确：由题意f(x2+x1)+f(x2-x1)=2f(x2)·f(x1)所以f(x1+x2)+f(x1-x2)=f(x2+x1)+f(x2-x1)所以f(x1-x2)=f(x2-x1)若x1-x2=x 则x2- 一道求导题目f(x1,x2)=x1^x2. 求二次方程的解#includestdio.h#includemath.hvoid main(){float a,b,c,p,q,x1,x2;scanf(%f,%f,%f ,&a,&b,&c);p=sqrt(b*b-4*a*c)/2;q=-b/2;x1=p+q;x2=p-q;printf(x1=%1.1f,x2=%1.1f ,x1,x2);} 杭电2001.运行结果是对的.为什么是wrong answer呢?#include#includemain(){float x1,x2,y1,y2;while(scanf(%f %f %f %f,&x1,&x2,&y1,&y2)!=EOF){printf(%.2f,sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));putchar(' ');}} 求C语言例题讲解：用迭代法求方程x=cos x的根,要求误差小于10的-6次方main(){double x1,x2;x1=0.0;x2=cos(x1);while(fabs(x2-x1)>le-6){x1=x2;x2=cos(x1);}printf(x=%f ,x2);}1、x1=0.0；x2=cos(x1);分别是什么意思2、while后 float x1,x2; x1=3/2; x2=x1/2; printf(%d%.1f,(int)x1,x2); 输出值是多少不是说强制转换类型是只限于本次运算吗?那计算x2的时候为什么不是用x1=3/2;而是用x1=1;? C语言写二次函数#include #include #include int main() { float a,b,c;float x1,x2,m;printf(input number a=:);scanf(%f,&a);printf(input number b=:);scanf(%f,&b);printf(input number c=:);scanf(%f,&c);m=b*b-4*a*c;x1=(-b+sqrt(m))/(2*a);x2 对于函数f(x)的定义域中任意的x1,x2(x1≠x2）,有如下结论1)f(x1+x2)=f(x1)*f(x2) (2)f(x1*x2)=f(x1)+f(x2) (3)[f(x1)-f(x2)]/(x1-x2)>0 (4) f[(x1+x2)/2]＞[f(x1)+f(x2)]/2 #include #include main() { float a,b,c,x1,x2,d; scanf(%f %f %f ,&a,&b,&c); d=b*#include#includemain(){float a,b,c,x1,x2,d;scanf(%f %f %f ,&a,&b,&c);d=b*b-4*a*c;if(d>=0 )if(d==0 ){ x1=x2=-b/(2*a);printf(x1=x2=%f ,x1);}else{ x1=(-b+sqrt(d))/(2