Question

срочного пожалуйста помогите ​

Answer 1

Пояснення:

1)

$\displaystyle\\(x^2-4x)(x^2-4x+1)=30\\\\x^2-4x=t\ \ \ \ \Rightarrow\\\\t*(t+1)=30\\\\t^2+t-30=0\\\\t^2+6x-5x-30=0\\\\t*(t+6)-5*(x-6)=0\\\\(t+6)*(t-5)=0\\\\t+6=0\\\\t_1=x^2-4x=-6\\\\x^2-4x+6=0\\\\D=4^2-4*1*6=16-24=-8.\\\\x\in\varnothing.\\\\t_2=x^2-4x=5\\\\x^2-4x-5=0\\\\x^2-5x+x-5=0\\\\x*(x-5)+(x-5)=0\\\\(x-5)*(x+1)=0\\\\x-5=0\\\\\x_1=5.\\\\x+1=0\\\\x_2=-1.$

Відповідь: х₁=5,   х₂=-1.

2)

$\displaystyle\\9x^4-10x^2+1=0\\\\x^2=t\geq 0\ \ \ \ \ \ \Rightarrow\\\\9t^2-10t+1=0\\\\9t^2-9t-t+1=0\\\\9t*(t-1)-(t-1)=0\\\\(t-1)*(9t-1)=0\\\\t-1=0\\\\t_1=x^2=1\\\\x_{1,2}=б1.\\\\9t-1=0\\\\9t=1\ |:9\\\\t=x^2=\frac{1}{9} \\\\x_{3,4}=б\frac{1}{3} .$

Відповідь: x₁=-1,   x₂=1,   x₃=-1/3,   x₄=1/3.

*)

$(x+2)^2(x^2+4x)=45\\\\(x+2)^2(x^2+4x+4-4)=45\\\\(x+2)^2((x+2)^2-4)=45\\\\(x+2)^2=t\geq 0\ \ \ \ \ \ \Rightarrow\\\\t*(t-4)=45\\\\t^2-4t-45=0\\\\t^2-9t+5t-45=0\\\\t*(t-9)+5*(t-9)=0\\\\(t-9)*(t+5)=0\\\\t-9=0\\\\t_1=(x+2)^2=9\\\\(x+2)^2-3^2=0\\\\(x+2+3)*(x+2-3)=0\\\\(x+5)*(x-1)=0\\\\x+5=0\\\\x_1=-5.\\\\x-1=0\\\\x_2=1.\\\\t+5=0\\\\t_2=(x+2)&^2=-5\notin.$

Відповідь: x₁=-5,  x₂=1.