[tex]\frac{ x(2x-1) }{3} =\frac{ 4x^2+7 }{6}\ \ |*6\\\\2x(2x-1)=4x^2+7\\\\4x^2-2x=4x^2+7\\\\4x^2-2x-4x^2=7\\\\-2x=7\ \ |:(-2)\\\\x=-\frac{7}{2}\\\\x=-3\frac{1}{2}[/tex]
[tex]x(2x-5) -1 =2x^2+7x-7\\\\ 2x^2-5x-1=2x^2+7x-7\\\\2x^2-5x-2x^2-7x=-7+1\\\\-12x=-6\ \ |:(-12)\\\\x=\frac{6}{12}\\\\x=\frac{1}{2}[/tex]
[tex]-3\frac{1}{2}+\frac{1}{2}=-3\\\\odp.\ Suma\ rozwiazan\ tych\ rownan\ jest\ liczba\ calkowita.[/tex]
