设:√(x²+5x+2)=t,则原方程可化为:
3√(x²+5x+2)]²+2√(x²+5x+2)=56
3t²+2t=56
3t²+2t-56=0
(3t+14)(t-4)=0
因t>0,则t=4
则:√(x²+5x+2)=4
x²+5x+2=16
x²+5x-14=0
(x+7)(x-2)=0
得:x=2或x=-7
设:√(x²+5x+2)=t,则原方程可化为:
3√(x²+5x+2)]²+2√(x²+5x+2)=56
3t²+2t=56
3t²+2t-56=0
(3t+14)(t-4)=0
因t>0,则t=4
则:√(x²+5x+2)=4
x²+5x+2=16
x²+5x-14=0
(x+7)(x-2)=0
得:x=2或x=-7