方程求解系统(最新版).doc

#include #include#includefloat delte,gh,a1,a2,b1,b2,c1,c2,d1,n;using namespace std; int MinY(int kb,int ke) int min, max; int r; max=kbke?kb:ke; min=kbke?kb:ke; if (max%min=0) return max; while(max%min!=0) r=max%min; max=min; min=r; return kb*ke/min; int main() float x1ans1,x1ans2,x1a,x1b,x1c;cout*nendl;cout 欢迎进入JWC方程求解系统nendl;cout*nendl;cout1.一元一次方程endl;cout对于 ax+b=c,; coutx1a; coutx1b; coutx1c;x1ans1=x1c-x1b; coutendl x1ax+x1b=x1cendl;cout解： x1ax=;printf(%.1fn,x1ans1); x1ans2=x1ans1/x1a;cout x=;printf(%.1fn,x1ans2);coutendlendlendl;cout2.一元二次方程endl;float x2ans1,x2ans2,a,b,c,k=0;cout对于 ax2+bx+c=0,endl; couta; coutb; coutc;if (c=0) k=2;x2ans1=b*b-4*a*c;coutendl ax2+bx+c=0endl;cout a=a ,b=b ,c=cendl;delte=x2ans1;cout0) cout0endl;gh=sqrt(delte);a1=0-b+gh;a2=0-b-gh;d1=2*a;c1=a1/d1;c2=a2/d1; cout x1=; printf(%.2fn,c1); cout x2=; printf(%.2fn,c2);if (delte=0) cout =0endl;gh=sqrt(delte);a1=0-b;d1=2*a;c1=a1/d1; cout x=; printf(%.2fn,c1);if (delte0) cout 0endl; cout 此方程无实数根！endl;coutendlendl; cout3.二元一次方程(整数)endl; float ka,kb,kc,kd,ke,kf,xy,xy8,xy9,ans1,ans2,ans3,ans4,q,p; int y1,y2,y3;cout对于 ax+by=c,endl; cout dx+ey=fendl; coutka; coutkb;coutkc; coutkd;coutke; coutkf; coutendl kax+kby=kc-endl;cout kdx+key=kf-endlendl; xy=MinY(kb,ke); y1=xy/kb; y2=xy/ke;ka=ka*y1; kb=kb*y1; kc=kc*y1;kd=kd*y2; ke=ke*y2; kf=kf*y2;cout解： *y1,得： kax+kby=kc-endl;cout *y2,得： kdx+key=kf-endl;ans1=kd-ka; ans2=kf-kc;cout -,得： ans1x=ans2-endl;ans4=ans2/ans1;cout /ans1,得： x=;printf(%.2f,ans4);cout-endl;cout 代入,得： ;q=ans4*ka;printf(%.2f,q);cout+kby=kc-endl; p=(kc-q)/kb; cout 化简,得： y=;printf(%.2f,p);coutendl 即： x=;printf(%.2f,ans4); coutendl y=;printf(%.2f,p); coutendlendlendl; cout4.分式方程(整数) endl;float ma,mb,mc,md,me,delte2,p1,p2,p3,kkkk;cout对于 b/(a+x)+c=e/(x+d),endl; coutma; coutmb;coutmc; coutmd;coutme; coutendl mb/(ma+x)+mc=me/(x+md)endl;cout解： 方程两边同乘(x+ma)(x+md),得endl;cout mb*(x+md)+mc*(x+ma)*(x+md)=(x+ma)*meendl;p1=mb+ma*mc+mc*md-me;p2=mb*md+ma*mc*md-ma*me; cout 即 mcx2+p1x+p2=0; cout a=mc ,b=p1 ,c=p2endl; if (mc!=0) p3=p1*p1-4*mc*p2;delte2=p3;float gh2,ma1,ma2,md1,mc1,mc2;cout0) cout0endl;gh2=sqrt(delte2);ma1=0-p1+gh2;ma2=0-p1-gh2;md1=2*mc;mc1=ma1/md1;mc2=ma2/md1;if (mc1+ma=0|mc1+md=0)coutx1=;printf(%.2f,mc1);if (mc2+ma=0|mc2+md=0)coutx2=;printf(%.2f,mc2);cout 检验：当x=; printf(%.2f,mc1); cout或; printf(%.2f,mc2); cout时，(x+ma)(x+md)=0，因此x=; printf(%.2f,mc1); cout或; printf(%.2f,mc2); cout不是原方程的解.endl; cout 所以，原分式方程无解.endl; if (mc2+ma!=0|mc2+md!=0) cout x2=; printf(%.2fn,mc2); cout 检验：当x=; printf(%.2f,mc1); cout时，(x+ma)(x+md)=0，因此x=; printf(%.2f,mc1); cout不是原方程的解；endl; cout 当x=; printf(%.2f,mc2); cout时，(x+ma)(x+md)0; cout 所以，原分式方程的解为x=; printf(%.2fn,mc2); if (mc1+ma!=0|mc1+md!=0) cout x1=; printf(%.2fn,mc1);if (mc2+ma!=0|mc2+md!=0) cout x2=; printf(%.2fn,mc2); cout 检验：当x=; printf(%.2f,mc1); cout或; printf(%.2f,mc2);cout时，(x+ma)(x+md)0endl; cout 所以，原分式方程的解为x1=; printf(%.2f,mc1); cout，x2=; printf(%.2fn,mc2);if (mc2+ma=0|mc2+md=0) cout x2=; printf(%.2fn,mc2); cout 检验：当x=; printf(%.2f,mc1); cout时，(x+ma)(x+md)0，因此x=; printf(%.2f,mc1); cout 当x=; printf(%.2f,mc2); cout时，(x+ma)(x+md)0，因此x=; printf(%.2f,mc2); cout不是原方程的解.endl; cout 所以，原分式方程的解为x=; printf(%.2fn,mc1); if (delte2=0) cout =0endl;gh2=sqrt(delte2);ma1=0-p1;md1=2*mc;mc1=ma1/md1;if (mc1+ma=0|mc1+md=0)cout x=;printf(%.2fn,mc1);cout 检验：当x=; printf(%.2f,mc1);cout时，(x+ma)(x+md)=0，因此x=;printf(%.2f,mc1);cout不是原方程的解.endl; cout 所以，原分式方程无解.endl; if (mc1+ma!=0|mc1+md!=0)cout x=; printf(%.2fn,mc1); cout 检验：当x=; printf(%.2f,mc1);cout时，(x+ma)(x+md)0;cout 所以，原分式方程的解为x=; printf(%.2fn,mc1); if (delte20) cout 0endl; cout 此分式方程无解！endl; if (mc=0) cout 即 p1x=p2*-1endl; cout 解得 x=; float ww1,ww2,ww3;ww1=p2*-1;ww2=ww1/p1;if (ww2+ma=0|ww2+md=0)cout 检验：当x=; printf(%.2f,ww2);cout时，(x+ma)(x+md)=0，因此x=