C语言解X3+AX2+B=0 一元三次的程序main(){Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;scanf("%f",&l);A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);B=73000*g2*g2*l*l/24;printf("%f,%f",A,B);x0=1.0;do{dx=(x0*x0*x0-
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![C语言解X3+AX2+B=0 一元三次的程序main(){Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;scanf(](/uploads/image/z/14811379-43-9.jpg?t=C%E8%AF%AD%E8%A8%80%E8%A7%A3X3%2BAX2%2BB%3D0+%E4%B8%80%E5%85%83%E4%B8%89%E6%AC%A1%E7%9A%84%E7%A8%8B%E5%BA%8Fmain%28%29%7BFloat+l%2Cg1%3D0.03275%2Cf%2Cg2%3D0.06544%2CA%2CB%2Ct1%3D-5%2Ct2%3D-5%2Cc%3D68.51%2Cx%2Cx1%2Cx0%2Cdx%3Bscanf%28%22%25f%22%2C%26l%29%3BA%3Dc-%2873000%2Ag1%2Ag1%2Al%2Al%29%2F%2824%2Ac%2Ac%29-73000%2A0.0000196%2A%28t2-t1%29%3BB%3D73000%2Ag2%2Ag2%2Al%2Al%2F24%3Bprintf%28%22%25f%2C%25f%22%2CA%2CB%29%3Bx0%3D1.0%3Bdo%7Bdx%3D%28x0%2Ax0%2Ax0-)
C语言解X3+AX2+B=0 一元三次的程序main(){Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;scanf("%f",&l);A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);B=73000*g2*g2*l*l/24;printf("%f,%f",A,B);x0=1.0;do{dx=(x0*x0*x0-
C语言解X3+AX2+B=0 一元三次的程序
main()
{
Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;
scanf("%f",&l);
A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);
B=73000*g2*g2*l*l/24;
printf("%f,%f",A,B);
x0=1.0;
do{
dx=(x0*x0*x0-A*x0*x0-B)/(3*x0*x0-2*A*x0);
x1=x0;
x0=x0-dx;
}
while(x0!=x1);
printf("\nc=%f\n",x0);
f=(g2*l*l)/(8*x0);
printf("f=%f",f);
}
问题出在哪里,求改正,写出正确的代码
C语言解X3+AX2+B=0 一元三次的程序main(){Float l,g1=0.03275,f,g2=0.06544,A,B,t1=-5,t2=-5,c=68.51,x,x1,x0,dx;scanf("%f",&l);A=c-(73000*g1*g1*l*l)/(24*c*c)-73000*0.0000196*(t2-t1);B=73000*g2*g2*l*l/24;printf("%f,%f",A,B);x0=1.0;do{dx=(x0*x0*x0-
X^3+A*X^2+B=0?是这意思吗?