因为 (2x^2-1)-(x^2-3x-2)=(2x^2+2x+3)-(x^2-x+2)=3x^2-x+1.
设a^2=2x^2-1,b^2=x^2-3x-2,c^2=2x^2+2x+3,d=2x^2-x+2.则
a^2-b^2=c^2-d^2 (1)
a+b=c+d (2)
∵a+b=c+d=0无解,∴(1)/(2)得
a-b=c-d (3)
(2)+(3)得 a=c,即 √(2x^2-1)=√(2x^2+2x+3)
由此解得 x=-2.
(2)-(3)也可求出x=-2.
经检验x=-2是原方程的根.