Question

помогите срочно даю 20 балов

Answer 1

a) (y²-3y-5)(y²-3y+1)=-5

Пусть y²-3y = t, тогда

(t-5)(t+1) = -5

t²+t-5t-5+5=0

t²-4t=0

t(t-4)=0

t-4=0 ; t₂=0

t₁=4

Вернёмся к замене:

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

1. y²-3y=0

y(y-3)=0

y-3=0 ; y₂=0

y₁=3

2. y²-3y=4

y²-3y-4=0

D = (-3)²-4*1*(-4)=9+16=25=5²

$y_{1} = \frac{3+5}{2*1} = \frac{8}{2} = 4$

$y_{2} = \frac{3-5}{2*1} = - \frac{2}{2} = -1$

Ответ: y₁ =  ; y₂ =

б) $\frac{y}{y+2} + \frac{y+2}{y-2} = \frac{8}{y^{2}-4 }$

y≠2 ;y≠-2

$\frac{y*(y-2)}{y+2} + \frac{(y+2)*(y+2)}{y-2} - \frac{8}{y^{2}-4 } =0$

$\frac{y^{2}-2y+y^{2}+4y+4 -8 }{y^{2}-4 } =0$

$\frac{2y^{2} +2y-4 }{y^{2}-4 } =0$

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

$\frac{2(y-1)(y+2) }{(y-2)(y+2)}=0$

$\frac{2(y-1) }{y-2}=0$ | * (y-2)

2(y-1)=0

y-1=0

y=1

Отвeт: y=1