O’zbekiston Respublikasi Raqamli Texnologiyalar Vazirligi Muhammad al-Xorazmiy nomidagi

O’zbekiston Respublikasi Raqamli Texnologiyalar Vazirligi
Muhammad al-Xorazmiy nomidagi
Toshkent Axborot Texnologiyalari Universiteti
Algoritmlarni loyihalash fani fani bo’yicha

AMALIY ISH

Bajardi: CAL013-1-guruh talabasi

Komiljonova Gavharshodbegim
Tekshirdi: Ravshanov Shohjaxon
Toshkent-2023
30-variant
N=30
n1={30/3}+1=1
n2={30/5}+1=1
n3={30/7}+1=3

x³ + x² x – 3 = 0

1. 
   Tenglamaning yechim joylashgan oralig’ini toping. Vatarlar usuli yordamida ildizlarni e=10^-5 aniqlik bilan hisoblash algoritmini tuzing. Bunday aniqllika erishish uchun zarur qadamlar sonini aniqlang.
   

#include
#include
using namespace std;
float f(float x){
return pow(x,3)+(x*x)-x-3; // vatarlar usuli
}
float f1(float x){
return 3*x*x+2*x-1;
}
float f2(float x){
return 6*x+2;
}
int main() {
float a,b,eps;
cin>>a>>b>>eps;
float x[100];
if(f1(a)*f2(a)>0)
x[0]=a;
else x[0]=b;
b=a;
x[1]=x[0]-(f(x[0])*(b-x[0]))/(f(b)-f(x[0]));
int i=0;
while(fabs(x[i+1]-x[i])>eps){
i++;
x[i+1]=x[i]-(f(x[i])*(b-x[i]))/(f(b)-f(x[i]));
}
cout<<"x="<

Ushbu masalani Nyuton usuli orqali yechish algoritmini tuzing. Ikkala usulda ham yechimlarni oling. Ko’rib chiqilgan usullar samaradorligini taqqoslang.


Dastur kodi :
#include
#include
using namespace std;
float f(float x){
return pow(x,3)+(x*x)-(x)-; // urinmalar usuli
}
float f1(float x){
return 3*x*x+2*x-1;
}
float f2(float x){
return 6*x+2;
}
int main() {
float a,b,eps;
cin>>a>>b>>eps;
float x[100];
if(f(a)*f2(a)>0)
x[0]=a;
else x[0]=b;
x[1]=x[0]-f(x[0])/f1(x[0]);
int i=0;
while(fabs(x[i+1]-x[i])>eps){
i++;
x[i+1]=x[i]-f(x[i])/f1(x[i]);
}
cout<<"x="<

Chiziqli dasturlash masalasini hosil qiling:

.
15x₁+10x₂+5x₃ ≤ 130
10x₁+4x₂+12x₃ ≤ 134
4x₁+15x₂+10x₃ ≤ 157
L= 1800x₁+2000x₂+1500x₃ → max