Odpowiedź:
[tex]a)\\\\(x+3)(1-2x)+2x^2=0\\\\x-2x^2+3-6x+2x^2=0\\\\-5x+3=0\\\\-5x=-3\ \ |:(-5)\\\\x=\frac{3}{5}\\\\\\b)\\\\(x+4)(x-1)=x(x+2)+9x-1\\\\x^2-x+4x-4=x^2+2x+9x-1\\\\x^2+3x-4=x^2+11x-1\\\\x^2+3x-x^2-11x=-1+4\\\\-8x=3\ \ |:(-8)\\\\x=-\frac{3}{8}[/tex]
[tex]c)\\\\x^2-(x+3)(x-3)=9\\\\x^2-(x^2-9)=9\\\\x^2-x^2+9=9\\\\9=9\\\\R\'ownanie\ \ to\.zsamo\'sciowe\ \ ma\ \ niesko\'nczenie\ \ wiele\ \ rozwiaza\'n\\\\\\d)\\\\(x+2)(x-3)=x^2+4\\\\x^2-3x+2x-6=x^2+4\\\\x^2-x-x^2=4+6\\\\-x=10\ \ |\cdot(-1)\\\\x=-10[/tex]
