Question

Me ajudem na número 7. COM A RESPOSTA POR FAVOR!

Answer 1

Olá!

$\dfrac{(3x-2)^2}{6}=\dfrac{2x+3}{3}-\dfrac{x(1-5x)}{2}\Leftrightarrow\\ \\ \\ \Leftrightarrow \dfrac{(3x-2)^2}{6}=\dfrac{2(2x+3)-3x(1-5x)}{6}\Leftrightarrow \\ \\ \\ \Leftrightarrow (3x-2)^2=4x+6-3x+15x^2\Leftrightarrow\\ \\ \Leftrightarrow 9x^2-12x+4=15x^2+x+6\Leftrightarrow 6x^2+13x+2=0\Leftrightarrow \\ \\ \Leftrightarrow x=\dfrac{-13\pm \sqrt{13^2-4\cdot 6\cdot 2}}{12}=\dfrac{-13\pm \sqrt{169-48}}{12}=\dfrac{-13\pm 21}{12}=\\ \\ \\ = \dfrac{2}{3}\;\text{ou}\;\dfrac{-17}{6}.$

Bons estudos!