Question

如何解1/x方程

Answer 1

更新1:

What does "x
y <> 0" mean?

x^2 + 1/x^2 = y^3 + 1/y^3 ... (1) 4x = 5y => y = 4x/5 ... (2) x
y <> 0 (1) => x^2 - y^3 + 1/x^2 - 1/y^3 = 0 x^2 - y^3 + (y^3 - x^2)/x^2y^3 = 0 (x^2 - y^3)(1 - 1/x^2y^3) = 0 x^2 - y^3 = 0 or x^2y^3 = 1 x^2 - (4x/5)^3 = 0 or x^2(4x/5)^3 = 1 x^2(1 - 64x/125) = 0 or 64x^5/125 = 1 x = 0 (rejected) or x = 125/64 or x^5 = 125/64 x = 125/64 or x = (125/64)^(1/5) x = 125/64 => y = 25/16 x = (125/64)^(1/5) => y = [(125/64)(1024/3125)]^(1/5) = (16/25)^(1/5) 2009-11-01 12:37:20 补充： xy都不等于0
wolframalpha/input/?i=Solve[{X^2%2B1%2F%28X^2%29%3D%3DY^3%2B1%2F%28Y^3%29%2C4X%3D5Y}]