Question

3x в 6 +6x в 4 - x во 2 -2=0

Answer 1

$3x^6+6x^4-x^2-2=0\\3x^4(x^2+2)-(x^2+2)=0\\(x^2+2)(3x^4-1)=0\\x^2+2ne 0; ,; ; (x^2+2>0); \\3x^4-1=0; ; to ; ; (sqrt3x^2-1)(sqrt3x^2+1)=0\\sqrt3x^2+1ne 0; ; ,; ; (sqrt3x^2+1>0); ; ,\\sqrt3x^2-1=0; ; to ; ; (sqrt[4]3x-1)(sqrt[4]3x+1)=0\\x_1=frac{1}{sqrt[4]3}; ,; ; x_2=-frac{1}{sqrt[4]3}$