Question

Помогите срочно !!!!!!! даю 60 балов

Answer 1

Пояснення:

1)

$\displaystyle\\x^4-3x^2-4=0\\\\(x^2)^2-3*x^2-4=0$

Нехай x²=t≥0         ⇒

$t^2-3t-4=0\\\\t^2-4t+t-4=0\\\\t*(t-4)+(t-4)=0\\\\(t-4)(t+1)=0\\\\t-4=0\\\\t_1=x^2=4\\\\x^2-4=0\\\\\x^2-2^2=0\\\\(x+2)(x-2)=0\\\\x_1=-2\ \ \ \ x_2=2.\\\\t+1=0\\\\t_2=x^2=-1\notin.\ \ \ \ \ \ \Rightarrow$

Відповідь: х₁=-2,  х₂=2.

2)

$\displaystyle\\\frac{x^2}{x+2}=\frac{4}{x+2}$

ОДЗ: х+2≠0     х≠-2.

$\displaystyle\\\frac{x^2}{x+2}-\frac{4}{x+2} =0\\\\\frac{x^2-4}{x+2}=0\\\\\frac{(x+2)(x-2)}{x+2} =0\ \ \ \ x+2\neq 0\\\\x-2=0 \\\\x=2.$

Відповідь: х=2.

3)

$x^3+2x^2-3x=0\\\\x*(x^2+2x-3)=0\\\\x_1=0.\\\\x^2+2x-3=0\\\\x^2+3x-x-3=0\\\\x*(x+3)-(x+3)=0\\\\(x+3)*(x-1)=0\\\\x+3=0\\\\x_2=-3.\\\\x-1=0\\\\x_3=1.$

Відповідь: х₁=0,  х₂=-3,  х₃=1.