Odpowiedź:
[tex]\frac{(2x-3)}{(x+1)} =x-(\frac{5}{3})\\ \\\frac{2x-3}{x+1} =x-\frac{5}{3}\\\\x\neq -1\\\\\frac{2x-3}{x+1} -x+\frac{5}{3}=0\\\\\frac{3(2x-3)-3x*(x+1)+5(x+1)}{3(x+1)}=0\\\\\frac{6x-9-3x^{2}-3x+5x+5 }{3(x+1)} =0\\\\\frac{8x-4-3x^{2} }{3(x+1)} =0\\\\8x-4-3x^{2} =0\\\\-3x^{2} +8x-4=0\\\\3x^{2} -8x+4=0\\\\x=\frac{-(-8)+\sqrt{(-8)^{2}-4*3*4 } }{2*3}\\\\x=\frac{8+\sqrt{16-48} }{6}\\\\x=\frac{8+\sqrt{16} }{6} \\\\x=\frac{8+4}{6} \\\\x=\frac{8-4}{6} \\\\x=2\\\\x=\frac{2}{3} \\\\[/tex]
[tex]x\neq 1\\\\x_{1}= \frac{2}{3} \\\\x_{2} =2[/tex]
