Срочно помогитеее даю 20 балов

a) (y²+4y-1)(y²+4y+3)=12

Пусть y²+4y = t, тогда

(t-1)(t+3) = 12

t²+3t-t-3-12=0

t²+2t-15=0

D = 2²-4*1*(-15)=4+60=64=8²

$x_{1} = \frac{-2+8}{2*1} = \frac{6}{2} = 3$

$x_{2} = \frac{-2-8}{2*1} = - \frac{10}{2} = -5$

Вернёмся к замене:

$\left \{ {{y^{2} +4y=3} \atop {y^{2} +4y=-5}} \right.$

$\left \{ {{y^{2}+4y-3=0 } \atop {y^{2}+4y+5=0}} \right.$

1. y²+4y-3=0

D = 4²-4*1*(-3)=16+12=28 = (2√7)²

$x_{1} = \frac{-4+2\sqrt{7} }{2*1} = -2+\sqrt{7}$

$x_{2} = \frac{-4-2\sqrt{7} }{2*1} = -2-\sqrt{7}$

2. y²+4y+5=0

D = 4²-4*1*5=16-20 - нет корней

Ответ: y₁ = - 2 - √7 ; y₂ = - 2 + √7

б) $\frac{y}{y+5} + \frac{y+5}{y-5} = \frac{50}{y^{2}-25 }$

y≠5 ;y≠-5

$\frac{y*(y-5)}{(y+5)(y-5)} + \frac{(y+5)(y+5)}{(y-5)(y+5)} - \frac{50}{y^{2}-25 } = 0$

$\frac{y^{2}-5y+y^{2}+10y+25- 50}{y^{2}-25 } = 0$

$\frac{2y^{2} + 5y-25}{y^{2}-25 } = 0$

$\frac{2(y-2,5)(y+5)}{(y+5)(y-5)} = 0$

$\frac{2(y-2,5)}{y-5} =0$ | * (y-5)

2y-5=0

2y = 5

y= 5/2 = 2,5

Отвте: y = 2,5