用换元法解方程(x^2+5x)^2+10x^2+50x+24=0

(x^2+5x)^2+10x^2+50x+24=0
(x^2+5x)^2+10(x^2+5x)+24=0
设(x^2+5x)=y
即y^2+10y+24=0
(y+4)(y+6)=0
y1=-4,y2=-6
x^2+5x=-4或x^2+5x=-6
(x+1)(x+4)=0或(x+2)(x+3)=0
x1=-1,x2=-4,x3=-2,x4=-3
祝你学习天天向上,加油!