关于分数指数幂的和的运算方式如(2+5^1/2)^1/3+(2-5^1/2)^1/3,求该式子值我只能用解高次方程的方法解,请问有没有不用高次方程的解法?
2019-04-12
关于分数指数幂的和的运算方式
如(2+5^1/2)^1/3+(2-5^1/2)^1/3,求该式子值
我只能用解高次方程的方法解,请问有没有不用高次方程的解法?
优质解答
(2+√5)^(1/3)+(2-√5)^(1/3)=?
[(2+√5)^(1/3)+(2-√5)^(1/3)]³
=(2+√5)+3[(2+√5)²(2-√5)]^(1/3)+3[(2+√5)(2-√5)²]^(1/3)+2-√5
=4+3[(2+√5)(2-√5)(2+√5)]^(1/3)+3[(2+√5)(2-√5)(2-√5)]^(1/3)
=4-3(2+√5)^(1/3)-3(2-√5)^(1/3)
=4-3[(2+√5)^(1/3)+(2-√5)^(1/3)]
移项得[(2+√5)^(1/3)+(2-√5)^(1/3)]³+3[(2+√5)^(1/3)+(2-√5)^(1/3)]-4=0
设[(2+√5)^(1/3)+(2-√5)^(1/3)]=x,则有:
x³+3x-4=x³-1+3x-3=(x-1)(x²+x+1)+3(x-1)=(x-1)(x²+x+4)=0
由于x²+x+4=(x+1/2)²-1/4+4=(x+1/2)²+15/4≧15/4>0,故必有x-1=0,即必有x=1.
也就是有(2+√5)^(1/3)+(2-√5)^(1/3)=1.
(2+√5)^(1/3)+(2-√5)^(1/3)=?
[(2+√5)^(1/3)+(2-√5)^(1/3)]³
=(2+√5)+3[(2+√5)²(2-√5)]^(1/3)+3[(2+√5)(2-√5)²]^(1/3)+2-√5
=4+3[(2+√5)(2-√5)(2+√5)]^(1/3)+3[(2+√5)(2-√5)(2-√5)]^(1/3)
=4-3(2+√5)^(1/3)-3(2-√5)^(1/3)
=4-3[(2+√5)^(1/3)+(2-√5)^(1/3)]
移项得[(2+√5)^(1/3)+(2-√5)^(1/3)]³+3[(2+√5)^(1/3)+(2-√5)^(1/3)]-4=0
设[(2+√5)^(1/3)+(2-√5)^(1/3)]=x,则有:
x³+3x-4=x³-1+3x-3=(x-1)(x²+x+1)+3(x-1)=(x-1)(x²+x+4)=0
由于x²+x+4=(x+1/2)²-1/4+4=(x+1/2)²+15/4≧15/4>0,故必有x-1=0,即必有x=1.
也就是有(2+√5)^(1/3)+(2-√5)^(1/3)=1.