Verified answer

Объяснение:

1)

[tex]\displaystyle\\\frac{3x+3}{3x+2} +\frac{2x-6}{3x-2}= 2\\\\[/tex]

ОДЗ: 3х+2≠0   х≠-2/3     3х-2≠0     х≠2/3.

[tex]\displaystyle\\(3x+3)*(3x-2)+(2x-6)*(3x+2)=2*(3x+2)*(3x-2)\\\\9x^2+3x-6+6x^2-14x-12=2*(9x^2-4)\\\\15x^2-11x-18=18x^2-8\\\\3x^2+11x+10=0\\\\3x^2+6x+5x+10=0\\\\3x*(x+2)+5*(x+2)=0\\\\(x+2)*(3x+5)=0\\\\x+2=0\\\\x_1=-2.\\\\3x+5=0\\\\x_2=-\frac{5}{3}.[/tex]

Ответ: x₁=-2   x₂=-5/3.

2)

[tex]\displaystyle\\x^4-2x^2-8=0[/tex]

Пусть х²=t≥0        ⇒

[tex]t^2-2t-8=0\\\\t^2-4t+2t-8=0\\\\t*(t-4)+2*(t-4)=0\\\\(t-4)*(t+2)=0\\\\t-4=0\\\\t=x^2=4\\\\x_1=-2\ \ \ \ \ \ x_2=2.\\\\t+2=0\\\\t=x^2=-2\ \notin\ (t\geq 0)[/tex]

Ответ: x₁=-2   x₂=2.