X^4+16x^2-17<=08 класс, используя дискриминант

X^4 + 16X^2 - 17 ≤ 0 X^2 = A ;  A > 0A^2 + 16A - 17 ≤ 0D = 256 + 68 = 324 √ D = 18 A1 = ( - 16 + 18 ) : 2 = 1 ( > 0 )A2 = ( - 16 - 18 ) : 2 = - 17 ( < 0 )-------------------X^2 = 1 X1 ≤  1 X2 ≤ - 1 

[latex] x^{4} +16 x^{2} -17 leq 0 \ x^{2} =t \ t^{2} +16t-17=0 \ D=256+68=324 \ sqrt{D} =17 \ t _{1} = frac{-16-18}{2} = frac{-34}{2} =-17(<0) \ t _{2} = frac{-16+18}{2} = frac{2}{2} =1(>0) \ \ x^{2} =1 \ x^{2} geq 1 \ x geq 1 \ x leq -1[/latex]