Question

Помогите решить ПЖ!!! Срочно!! Даю 50 баллов!!!​

Answer 1

$\displaystyle\bf\\1)\\\\2x^{4} -10x^{2} +8=0 \ |:2\\\\x^{4} -5x^{2} +4=0\\\\x^{2}=m \ , \ m > 0\\\\m^{2} -5m+4=0\\\\D=(-5)^{2} -4\cdot 4=25-16=9=3^{2} \\\\\\m_{1}=\frac{5-3}{2} =1\\\\\\m_{2} =\frac{5+3}{2} =4\\\\\\1)\\\\x^{2} =1\\\\x_{1} =-1 \ \ , \ \ x_{2} =1\\\\2)\\\\x^{2} =4\\\\x_{3} =-2 \ \ , \ \ x_{4} =2\\\\Otvet \ : -1 \ ; \ 1 \ ; \ -2 \ , \ 2$

$\displaystyle\bf\\2)\\\\\frac{3}{y+2} -\frac{y-2}{y} =0\\\\\\\frac{3\cdot y-(y-2)\cdot(y+2)}{y(y+2)} =0\\\\\\\frac{3y-y^{2}+4 }{y(y+2)} =0\\\\\\\frac{y^{2} -3y-4}{y(y+2)} =0\\\\\\\left \{ {{y^{2} -3y-4=0} \atop {y\neq 0 \ , \ y\neq -2}} \right.\\\\\\y^{2} -3y-4=0\\\\Teorema \ Vieta:\\\\y_{1} + y_{2} =3\\\\y_{1} \cdot y_{2} =-4\\\\y_{1} =-1 \ \ ; \ \ y_{2} =4\\\\Otvet \ : -1 \ : \ 4$