Odpowiedź:
[tex]a)\\\\(x+1)(x-1)=(2x+5)(x-1)\\\\x^2-1=2x^2-2x+5x-5\\\\x^2-1=2x^2+3x-5\\\\x^2-1-2x^2-3x+5=0\\\\-x^2-3x+4=0\ \ |\cdot(-1)\\\\x^2+3x-4=0\\\\a=1\ \ ,\ \ b=3\ \ ,\ \ c=-4\\\\\Delta=b^2-4ac\\\\\Delta=3^2-4\cdot1\cdot(-4)=9+16=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-3-5}{2\cdot1}=\frac{-8}{2}=-4\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-3+5}{2\cdot1}=\frac{2}{2}=1[/tex]
[tex]b)\\\\x^2-2x=3\\\\x^2-2x-3=0\\\\a=1\ \ ,\ \ b=-2\ \ ,\ \ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=(-2)^2-4\cdot1\cdot(-3)=4+12=16\\\\\sqrt{\Delta}=\sqrt{16}=4\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2\cdot1}=\frac{2-4}{2}=\frac{-2}{2}=-1\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2\cdot1}=\frac{2+4}{2}=\frac{6}{2}=3[/tex]
[tex]c)\\\\3x^2-x=0\\\\x(3x-1)=0\\\\x=0\ \ \ \ \vee\ \ \ \ 3x-1=0\\\\x=0\ \ \ \ \vee\ \ \ \ 3x=1\ \ |:3\\\\x=0\ \ \ \ \vee\ \ \ \ x=\frac{1}{3}\\\\\\d)\\\\5x+2=0\\\\5x=-2\ \ |:5\\\\x=-\frac{2}{5}\\\\lub\ \ je\'sli\ \ taki\ \ zapis\\\\5x^2+2=0\\\\a=5\ \ ,\ \ b=0\ \ ,\ \ c=2\\\\\Delta=b^2-4ac\\\\\Delta=0^2-4\cdot5\cdot2=0-40=-40\\\\\Delta<0\ \ brak\ \ rozwiazania[/tex]
