已知x+x^(-1)=3,求下列各式的值(1)x^0.5+x^-0.5 (2)x^2+x^-2 (3)x^2-x^-2
已知x+x^(-1)=3,求下列各式的值(1)x^0.5+x^-0.5 (2)x^2+x^-2 (3)x^2-x^-2
按正常解法为:(1)x^0.5+x^-0.5
=√(x^0.5+x^-0.5)^2
=√[(x+x^(-1))+2]
=√5
(2)x^2+x^-2
=(x+x^-1)^2-2
=9-2=7
(3)x^2-x^-2
=(x+x^-1)(x-x^-1)
=3(x-x^-1)
x-x^-1=√[(x+x^-1)^2-4]=√5
或x-x^-1=-√[(x+x^-1)^2-4]=-√5
x^2-x^-2=3√5
或
x^2-x^-2=-3√5
(可是为什么第一问中y=√5而不是±√5,在第三问中却是±√5,有什么区别吗?)