Question

x⁴+2x³-2x²+2x-3=0

Answer 1

x^4 + 2x^3 - 2x^2 + 2x - 3 = ( x^4 - 2x^2 + 1) + 2( x^3 + x + 2)

                                    = ( x^2 - 1)^2 + 2 ( x^3 + x^2 - x^2 - x + 2x + 2)

                                   = ( x^2 - 1)^2 + 2 (x^2 - x + 2)(x+1)

                                  = (x +1)( ( x+1)(x-1)^2 + 2(x^2-x+2) )

                               = 0

TH1 : x+ 1 = 0 => x = -1

TH2: (x+1)(x-1)^2 + 2 (x^2 - x +2) = 0

<=> (x+1)(x-1)^2 + 2(x-1)x + 4 = 0

<=> (x-1) ( x^2 - 1 + 2x) + 4 = 0

<=> (x - 1) ( x-1) ^2 + 4 = 0

<=> (x-1)^3 = -4

<=> x - 1 = \(\sqrt[3]{-4}\)

<=> x = \(\sqrt[3]{-4}\)+ 1

Vậy x có 2 giá trị thỏa mãn