giả sử phương trình $x^3-x^2+ax+b=0$ có $3$ nghiệm phân btệt.  CM $a^2+2b>0$
giả sử pt có 3 no :x1,x2,x3.nên ta có:(x-x1)(x-x2)(x-x3) =0.khai triển ra:
                   x^{3} -(x1+x2+x3)x^{2} +(x1x2+x2x3+x1x3)x -x1x2x3=0
Đồng nhất vs pt ban đầu ta đc: x1+x2+x3=1, x1x2+x2x3+x1x3=a, x1x2x3=-b
Thay vào biểu thức:a2 +2b =(x1x2)2 +(x1x3)2+(x2x3)2 >0




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