Question

Решите срочно уравнение 5)

Answer 1

$x^4+3x^2=frac{1}{x^4+3x^2+2}; ; ,; ; ODZ:; ; x^4+3x^2+2ne 0\\t=x^4+3x^2geq 0; ; ,; ; ; t=frac{1}{t+2}; ,; ; frac{t^2+2t-1}{t+2}=0; ; ,; ; left { {{t^2+2t-1=0} atop {tne -2quad }} right.\\t^2+2t-1=0; ; ,; ; (t+1)^2-2=0; ; ,; ; (t+1-sqrt2)(t+1+sqrt2)=0; ,\\t_1=-1+sqrt2>0; ; ,; ; t_2=-1-sqrt2<0; ,\\a); ; x^4+3x^2=-1+sqrt2; ; ,; ; x^4+3x^2+1-sqrt2=0; ,$

$D=9-4cdot (1-sqrt2)=5+4sqrt2; ; ,\\x^2=frac{-3-sqrt{5+4sqrt2}}{2}<0; ; ne; podxodit\\x^2=frac{-3+sqrt{5+4sqrt2}}{2}approx 0,1322; ; to ; ; x_{1,2}=pm sqrt{frac{-3+sqrt{5+4sqrt2}}{2}}approx pm 0,3637$

$b); ; x^4+3x^2=-1-sqrt2<0; ; to ; ; xin varnothing ,; tak; kak; ; (x^4+3x^2)geq 0; .\\Otvet:; ; x_1=-sqrt{frac{-3+sqrt{5+4sqrt2}}{2}}; ; ,; ; x_2=sqrt{frac{-3+sqrt{5+4sqrt2}}{2}}}; .$