令y=x^1/2+(x+5)^1/2 (y>0)y^2=x+(x+5)+2[x(x+5)]^1/2原方程变为y+y^2-2x-5=25-2x即y^2+y=30,所以(y+6)(y-5)=0,y=5,即x^1/2+(x+5)^1/2=5；(x+5)^1/2=5-x^1/2两边平方,有x+5=25+x-10*x^1/2,所以x=4