Odpowiedź:
[tex]a)\\\\(x+3)(x+2)-x^2=0\\\\x^2+2x+3x+6-x^2=0\\\\5x+6=0\\\\5x=-6\ \ |:5\\\\x=-\frac{6}{5}\\\\x=-1\frac{1}{5}\\\\\\b)\\\\(x-4)(x+5)-x^2=3\\\\x^2+5x-4x-20-x^2=3\\\\x-20=3\\\\x=3+20\\\\x=23\\\\\\c)\\\\x^2-(x-3)(x-1)=1\\\\x^2-(x^2-x-3x+3)=1\\\\x^2-(x^2-4x+3)=1\\\\x^2-x^2+4x-3=1\\\\4x=1+3\\\\4x=4\ \ |:4\\\\x=1[/tex]
[tex]d)\\\\2x^2-(2x+1)(x-2)=5\\\\2x^2-(2x^2-4x+x-2)=5\\\\2x^2-(2x^2-3x-2)=5\\\\2x^2-2x^2+3x+2=5\\\\3x=5-2\\\\3x=3\ \ |:3\\\\x=1\\\\\\e)\\\\x^2-2=(x+2)(x+4)\\\\x^2-2=x^2+4x+2x+8\\\\x^2-2=x^2+6x+8\\\\x^2-x^2-6x=8+2\\\\-6x=10\ \ |:(-6)\\\\x=-\frac{10}{6}\\\\x=-1\frac{4}{6}\\\\x=-1\frac{2}{3}\\\\\\f)\\\\(x-3)(4x+2)=4x^2+3\\\\4x^2+2x-12x-6=4x^2+3\\\\4x^2-10x-4x^2=3+6\\\\-10x=9\ \ |:(-10)[/tex]
[tex]x=-\frac{9}{10}[/tex]
