Rozwiązanie w załączniku.
[latex]zad.1)\ \(x-2)^2=(x-1)(x+1)-3x\ \x^2-4x+4=x^2-1-3x\ \x^2-4x-x^2+3x =-1-4\ \-x=-5/*(-1)\ \x=5[/latex] [latex]zad.2)\ \x^2-5x+4=0\ \Delta = b^{2}-4ac = (-5)^{2}-4*1*4=25-16=9 \ \sqrt{Delta }=sqrt{9}=3\ \x_{1}=frac{-b-sqrt{Delta }}{2a} =frac{5-3}{2}=frac{2}{2}=1 \ \x_{2}=frac{-b+sqrt{Delta }}{2a} =frac{5+3}{2}=frac{8}{2}=4[/latex] [latex]zad.3) \ \2x^{2}-3x+1<0\ \Delta = b^{2}-4ac = (-3)^{2}-4*2*1=9-8=1\ \sqrt{Delta }=sqrt{1}=1\ \x_{1}=frac{-b-sqrt{Delta }}{2a} =frac{3-1}{4}=frac{2}{4}=frac{1}{2} \ \x_{2}=frac{-b+sqrt{Delta }}{2a} =frac{3+1}{4}=frac{4}{4}=1[/latex] [latex]a>0 \ \ramiona paraboli skierowane do gory\ \x in (frac{1}{2}) ,1 )[/latex] [latex]zad.4)\ \egin{cases}2x+y=3 /*(-3)\x+3y=4 end{cases}\ \egin{cases}-6x-3y=-9\x+3y=4 end{cases}\-------- \-5x=-5/:(-5)\ \x=1[/latex] [latex]2*1+y=3\ \y=3-2\ \y=1 \ \egin{cases}x=1\y=1 end{cases}[/latex] załącznik do zadania 3)
