zad.1. [latex]-x^2+6x+7 geq 0\ Delta=(6)^2-4*(-1)*7=36+28=64\ sqrt{Delta}=sqrt{64}=8\ x_1=frac{-b-sqrt{Delta}}{2a}=frac{-6-8}{-2}=frac{-14}{-2}=7\ x_2=frac{-b+sqrt{Delta}}{2a}=frac{-6+8}{-2}=frac{2}{-2}=-1[/latex] Teraz rysujesz wykres (załącznik) i zapisujesz odpowiedź x∈<-1;7> zad.2. [latex]2x^3+x^2-6x-3=0\ x^2(2x+1)-3(2x+1)=0\ (x^2-3)(2x+1)=0\ 1^o x^2-3=0\ (x-sqrt{3})(x+sqrt{3})=0\ x=sqrt{3} oraz x=x=-sqrt{3}\ 2^o2x+1=0\ 2x=-1\ x=-frac{1}{2}\ Odp. x=sqrt{3} , x=-sqrt{3}, x=-frac{1}{2}[/latex] zad.3. a) [latex]|x+3|>4\ x+3>4 i x+3<-4\ x>1 i x<-7\ [/latex] x∈(-∞;-7) ∨ (1;+∞) b) [latex]|x-5| leq 6\ x-5 leq 6 oraz x-5 geq -6\ x leq 11 oraz x geq -1[/latex] x∈<-1;11> zad.4. [latex]sin alpha =frac{3}{4}\ sin^2 alpha +cos^2 alpha=1\ cos alpha =sqrt{1-sin^2 alpha } \ cos alpha =sqrt{1-(frac{3}{4})^2}=sqrt{1-frac{9}{16}}=sqrt{frac{7}{16}}=frac{sqrt{7}}{4}\ \ ctg alpha =frac{cos alpha }{sin alpha }=frac{frac{sqrt{7}}{4}}{frac{3}{4}}=frac{sqrt{7}}{3}\ \ 3-ctg^2 alpha =3-(frac{sqrt{7}}{3})^2=3-frac{7}{9}=2frac{2}{9}[/latex]
