Odpowiedź:
a)
[tex]-x^{2}+2+3\geq 0\\ \\-x^{2} +5\geq 0\\\\x^{2} -5\leq 0\\\\x^{2} \leq 5\\\\|x|\leq \sqrt{5}\\\\[/tex]
x ∈ [tex][-\sqrt{5},\sqrt{5}][/tex]
b)
[tex]3x^{2} +6x+10=0\\\\[/tex]
x = -6±[tex]\sqrt{6^{2}-4*3*10 }[/tex] / 2+3
[tex]3*0^{2} +6*0+10=10[/tex]
x ∈ R
c)
-2x^2+8x−8≥0
2x^2−8x+8≤0
2x^2−8x+8=0
x = 8±0/4
x=2
d)
[tex]2x-x*(x+3<(x-1)^{2} -2\\\\x_{1} =-\frac{1}{2} \\\\x_{2} =1\\\\x<-\frac{1}{2} \\\\-\frac{1}{2} 1\\\\x_{1}=-1\\ \\x_{2}=0\\ \\x_{3}=2\\ \\[/tex]
x<-1/2 lub x>1 są rozwiązaniami. Pozdrawiam.
