Question

初中代数题

已知x=2+√3，y=2-√3，计算代数式（x+y/x-y - x-y/x+y）·（1/x² - 1/y平方）

Answer 1

xy=2²-(√3)²=1

原式={[(x+y)²-(x-y)²]/(x+y)(x-y)]}[(y²-x²)/x²y²]
=[4xy/(x²-y²)][(y²-x²)/x²y²]
=-4/(xy)
=-4

Answer 2

由题意得：
x+y=2+√3+2-√3=4
x-y=2+√3-(2-√3)
=2√3 xy
=(2+√3)(2-√3)
=1

[(x+y)/(x-y)-(x-y)/(x+y)] ×(1/x^2-1/y^2)
=[4/(2√3) - (2√3)/4]×[-(x^2 - y^2)/(x^2y^2)]
=[(√3)/6]×[-(x+y)(x-y)/(xy)^2]
=[(√3)/6]×(-8√3)
=-4