Matlab va Mathcadda elementar matematik masalalarni yechish 1-topshiriq. Quydagi ifodalarni Matlabda yozing va natija oling: Ma’lum bir tenglik algebraik o’zgaruvchilarga bog’liq bo’la uning shu o’zgaruvchilaridagi qiymatini topishning ikki usuli mayjud birinchi usul o’zgaruvchilarni elon qilib algebraik tenglamani avval kiritib keyin o’zgaruvchilarni malum bir uzgarmaslarga tenglash orqali, ikki ucgun o’zgaruvchilar borligini elon qilmiy usha algebraic tenglamani o’zgaruvchilarini malum bir uzgarmaslarga tenglab keyin tenglikni kiritish mumkin. a=3.56; b=exp(0.314); y=(log(b+1.4)/log(a))^(-3/4) y = 1.1799 a=3*pi/7;b=83*pi/180; abs(log(1+sin(a)+log(1-sin(b)/log(exp(1))))) ans = 4.2185 a=5*pi/7;b=0.3*pi; (sin(a+b)^2-sin(a-b)^2)^(1/3) ans = 0.4876 + 0.8445i p=log(3)/log(exp(1));q=log(3); 3*(p^(-2/3)+q^(-1/2))*(p*q)^(1/3) ans = 6.0474 m=3.6485; 2/3*m*sqrt(m*(m*(m)^(1/4))^(1/3)) ans = 6.0840

Expand orqali algebraic ifodalardagi qavislarni ochish mumkin quyida ushbu buyroq orqali qavslarni ochib chiqganmiz. 2-topshiriq. expand va simplify komandalaridan foydalanib, ifodani soddalashtiring EXPAND() syms x Y=(x-4)*(x-15.4)*(x+10.3); expand(Y) ans = x^3 - (91*x^2)/10 - (6911*x)/50 + 15862/25 expand((x-2)*(x-15)*(x+103)) ans = x^3 + 86*x^2 - 1721*x + 3090 expand((x-2.4)*(x-1.5)*(x+1.03) ) ans = x^3 - (287*x^2)/100 - (417*x)/1000 + 927/250 expand((x-4.8)*(x-1.75)*(x+13) ) ans = x^3 + (129*x^2)/20 - (307*x)/4 + 546/5 expand((x-8.2)*(x-2.75)*(x+1.3)) ans = x^3 - (193*x^2)/20 + (1663*x)/200 + 5863/200

Simplify buyrug’i orqali esa ifodani soddalashtirishimiz mumkin ushbu buyruqdan foydalanishni quyida ko’rsatilgan. simplify() syms x Y=(x-4)*(x-15.4)*(x+10.3); simplify(Y) ans = (x - 4)*(x - 77/5)*(x + 103/10) simplify( (x-2)*(x-15)*(x+103) ) ans = (x - 2)*(x - 15)*(x + 103) simplify((x-2.4)*(x-1.5)*(x+1.03)) ans = (x - 3/2)*(x - 12/5)*(x + 103/100) simplify( (x-4.8)*(x-1.75)*(x+13) ) ans = (x - 7/4)*(x + 13)*(x - 24/5) simplify((x-8.2)*(x-2.75)*(x+1.3) ) ans = (x - 11/4)*(x + 13/10)*(x - 41/5)

Ifodalarni ko’paytiruvchilarga ajratish uchun factor buyrug’idan foydalanishimiz mumkin bu bizga ko’plab qulayliklar tug’diradi ushbu buyruqdan quyidagicha foydalanamiz. 3-topshiriq. Ifodani ko’paytuvchilarga ajrating syms x Y=x^3-32/5*x^2-23/5*x+12; factor(Y) ans = [1/5, 5*x^3 - 32*x^2 - 23*x + 60] factor(x^3+13/2*x^2-26*x-315/2) ans = [1/2, x + 7, x - 5, 2*x + 9] factor(x^3-17/10*x^2-69/10*x+9) ans = [1/10, 2*x + 5, x - 3, 5*x - 6] factor(x^3-16/5*x^2-92/5*x+208/5) ans = [1/5, 5*x - 26, x - 2, x + 4] factor(x^3+27/10*x^2-26/5*x-42/5) ans = [1/10, 5*x + 6, 2*x + 7, x - 2]