Question

Решить уравнение(в файле):

Answer 1

$\displaystyle\bf\\\frac{y+2}{8y^{3} +1} -\frac{1}{4y+2}=\frac{y+3}{8y^{2} -4y+2} \\\\\\\frac{y+2}{(2y+1)(4y^{2} -2y+1)} -\frac{1}{2\cdot(2y+1)} -\frac{y+3}{2(4y^{2} -2y+1)} =0\\\\\\\frac{2\cdot(y+2)-(4y^{2} -2y+1)-(y+3)\cdot(2y+1)}{2(2y+1)(4y^{2} -2y+1)} =0\\\\\\\frac{2y+4-4y^{2} +2y-1-2y^{2}-y-6y-3 }{2(2y+1)(4y^{2}-2y+1) }=0\\\\\\\frac{-6y^{2} -3y}{(2y+1)(4y^{2}-2y+1) } =0$

$\displaystyle\bf\\\frac{2y^{2}+y }{(2y+1)(4y^{2} -2y+1)} =0\\\\\\\frac{y(2y+1)}{(2y+1)(4y^{2} -2y+1)}=0\\\\\\\frac{y}{4y^{2} -2y+1}=0\\\\\\y=0 \ \ \ ; \ \ \ y\neq -0,5$