Question

alguém pode me ajudar?? !! pfvr .. $\frac{x}{3} + \frac{x}{2} = \frac{ 6x^{2}+1 }{6}$

Answer 1

$\dfrac{x}{3} + \dfrac{x}{2} = \dfrac{ 6x^{2}+1 }{6}\to~~~~mmc(3,2,6)=6\\\\ 2x+3x=6x^2+1\to\\\\ 6x^2+1-2x-3x=0\to\\\\ 6x^2-5x+1=0\\\\a=6~~;~~b=-5~~;~~c=1\\\\\\ \Delta=b^2-4ac\to \Delta=(-5)^2-4.6.1\to \Delta=25-24\to \boxed{\Delta=1}$



$x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \to~~~ x= \dfrac{-(-5)\pm \sqrt{1} }{2.6} \to~~~ x= \dfrac{5\pm 1 }{12} \to\\\\\\ x'= \dfrac{5+ 1 }{12} \to~~~ x'= \dfrac{6 }{12} \to~~~ \boxed{~x'= \dfrac{1 }{2} ~}\\\\\\ x''= \dfrac{5- 1 }{12} \to~~~ x''= \dfrac{4 }{12} \to~~~ \boxed{~x''= \dfrac{1 }{3} ~}\\\\\\\\ \Large\boxed{\boxed{ S=\left\{ \dfrac{1 }{3} ~~;~~ \dfrac{1 }{2} \right\} }}}$

Answer 2

Taammyy,
Vamos passo a passo

            $\frac{x}{3} + \frac{x}{2} =\frac{6x^2+1}{6} \\ \\ \frac{2x+3x}{6} = \frac{6x^2+1}{6} \\ \\ 5x=6x^2+1 \\ \\ 6x^2-5x+1=0$
 
Fatorando
                     (2x - 1)(3x - 1) = 0
Cada fator deve ser nulo    
                    2x - 1 = 0
                            2x = 1
                                $x1= \frac{1}{2}$             
                    3x - 1 = 0
                            3x = 1
                                $x2 = \frac{1}{3}$
 
                                              S = {$\frac{1}{3} , \frac{1}{2}$}