Odpowiedź:
[tex](x-2)^2+x=(x+4)^2-7\\\\x^2-4x+4+x=x^2+8x+16-7\\\\x^2-3x+4=x^2+8x+9\\\\x^2-3x-x^2-8x=9-4\\\\-11x=5\ \ /:(-11)\\\\x=-\frac{5}{11}[/tex]
[tex](3x-1)^2-4<(3x+2)(3x-2)\\\\9x^2-6x+1-4<9x^2-4\\\\9x^2-6x-3<9x^2-4\\\\9x^2-6x-9x^2<-4+3\\\\-6x<-1\ \ /:(-6)\\\\x>\frac{1}{6}\\\\x\in(\frac{1}{6},+\infty)[/tex]
