已知:根号x=根号a+[1/(根号a)] ,(0已知:根号x=根号a+[1/(根号a)] ,(0求代数式[(x²+x-6)/x]÷[(x+3)/(x²-2x)]-[(x-2)+根号(x²-4x)]/[x-2-根号(x²-4x)]的值.

问题描述:

已知:根号x=根号a+[1/(根号a)] ,(0已知:根号x=根号a+[1/(根号a)] ,(0求代数式[(x²+x-6)/x]÷[(x+3)/(x²-2x)]-[(x-2)+根号(x²-4x)]/[x-2-根号(x²-4x)]的值.

根号x=根号a+[1/(根号a)],两边都平方,得x=a + 1/a + 2
[(x²+x-6)/x]÷[(x+3)/(x²-2x)] = [(x+3)(x-2)/x] * [x(x-2)/(x+3)]=(x-2)²=(a+1/a)²
x²-4x=(x-2)²-4=(a+1/a)² -4 =(a- 1/a)²
由于0