Помогите решить пожалуйста

Ответ:

1.x1 = 2, x2 = -2, x3 = -√2 х4 = √2

2. $\frac{6(x-2,5)(x-\frac{1}{3} ) }{3(x-\frac{1}{3}) }$ = $\frac{6x-15}{3}$

Объяснение:

x^4 - 6x^2 + 8 = 0

x^2 = y

y^2 - 6y + 8 = 0

D = (-6)^2 - 4(1 * 8) = 36 - 32 = 4

y1 = -(-6) - √4 / 2 * 1 = 6-2 / 2 = 4/2 = 2

y2 = -(-6)+√4/2*1 = 6+2/2 = 8/2 = 4

x^2 = y

x1 = √2

x2 = -√2

x3 = -2

x4 = 2

2. 6x^2+13x-5/3x-1

$\left \{ {{6x^2+13x-5=0} \atop {3x-1\neq0}} \right.$

$\left \{ {{6x^2+13x-5=0} \atop {x\neq \frac{1}{3} }} \right.$

$6x^{2}$ +13x-5=0

D= $13^{2}$ -4(6*(-5))=169+120=289= $17^{2}$

$x_{1}$ = $\frac{-13-\sqrt{289} }{2*6}$ = $\frac{-30}{12}$ =-2,5

$x_{2}$ = $\frac{-13+\sqrt{289} }{2*16}$ = $\frac{4}{12}$ = $\frac{1}{3}$

$\frac{6(x-2,5)(x-\frac{1}{3} ) }{3(x-\frac{1}{3}) }$ = $\frac{6x-15}{3}$ (х- $\frac{1}{3}$ сокращается, и раскрываем скобки)