Question

ДАЮ 90 баллов; Найдите разность между большим и меньшим корнями уравнения $\frac{4}{y+3} - \frac{1}{y-3} = \frac{5}{3-y} - 1$

Answer 1

$\displaystyle\bf\\\frac{4}{y+3} -\frac{1}{y-3} =\frac{5}{3-y} -1\\\\\\\frac{4}{y+3} -\frac{1}{y-3} +\frac{5}{y-3} +1=0\\\\\\\frac{4}{y+3} +\frac{4}{y-3} +1=0\\\\\\\frac{4\cdot(y-3)+4\cdot(y+3)+(y-3)(y+3)}{(y+3)(y-3)} =0\\\\\\\frac{4y-12+4y+12+y^{2} -9}{(y+3)(y-3)} =0\\\\\\\frac{y^{2} +8y-9}{(y+3)(y-3)} =0\\\\\\\left \{ {{y^{2} +8y-9=0} \atop {y+3\neq 0 \ \ ; \ \ y-3\neq 0}} \right. \\\\\\\left \{ {{(y-1)(y+9)=0} \atop {y\neq -3 \ \ ; \ \ y\neq 3}} \right.$

$\displaystyle\bf\\\left \{ {{\left[\begin{array}{ccc}y_{1} =1\\y_{2}=-9 \end{array}\right } \atop {y\neq -3 \ \ ; \ \ y\neq 3}} \right. \\\\1-(-9)=1+9=10\\\\Otvet:10$