Zadanie 1 [latex]-9(2m-3)+(m-3)^3-(m+2)(m-2)-m^3\\ =-18m+27+m^3-3*m^2*3+3*m*3^2-3^3+\ -(m^2-4)=\\ = -18m+27+m^3-3*m^2*3+3*m*3^2-3^3+\ -m^2-m^3+4=\\= -18m+27+m^3-9m^2+27m-27-m^3+4-m^2=\ =m^3-m^3-9m^2-m^2-18m+27m+27-27+4=\ =-10m^2+9m+4[/latex] Zadanie 2 Mianownik musi być różny od zera więc [latex]x^2-3 eq0 [/latex] Δ=0-4*1*(-3)=12 √Δ=√(12)=√(4*3)=2√3 [latex]x_1=frac{0-2sqrt3}{2*1} =frac{-2sqrt3}{2}=-sqrt3 \\ x_2=frac{0+2sqrt3}{2*1} =frac{2sqrt3}{2}=sqrt3[/latex] Dziedzina wychodzi nam D=R{-√3, √3} Sprowadzenie do najprostrzej postaci: [latex]frac{3x^4-27}{x^2-3}=frac{3(x^2-3)(x^2+3)}{x^2-3}=3(x^2+3)=3x^2+9[/latex]
[latex]\1. \-9(2m-3)+m^3-9m^2+27m-27-(m^2-4)-m^3= \-18m+27+m^3-9m^2+27m-27-m^2+4-m^3= \-10m^2+9m+4 \2. \frac{3x^4-27}{x^2-3}=frac{3(x^4-9)}{x^2-3}=frac{3(x^2+3)(x^2-3)}{x^2-3}= \3(x^2+3)=3x^2+9 \zal. \x^2-3 eq0 \(x+sqrt3)(x-sqrt3) eq0 \x eq-sqrt3 i x eqsqrt3 \D=Rackslash{-sqrt3,sqrt3}[/latex]
