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他的回答已经明确这个多项式怎么在实数域或者复数域上分解了，我简单提一下怎么在有理数域或者整数环上分解。 由于这个多项式 x^n-y^n 是齐次的，所以我们只需要考虑 f (x) = x^n-1 这个 …

Write out your expanded expression in two rows multiplying first by x to get the first row and then by − y to get the second row. If you write the second row so that expressions with the same …

2009年9月27日 · E.g. x*x^(n-2)*y cancels y*x^(n-1), x*x^(n-3)*y^2 cancels y*x^(n-2)*y. I know you can't write out all of the terms. You'll have to use the '...' to express what you mean.

x^n-y^n =(x-y)[x^(n-1)+x^(n-2)y+x^(n-3)y^2+...+x^2y(n-3)+xy^(n-2)+y^(n-1)] 注意：即中括号内每项次数和为n-1，将x-y乘进去，即可直接验证。 如果是加的话， n为奇数时，可直接将 …

I am looking for a general expansion of $x^{n}-y^{n}$ with $x,y>0$ and $n$ being real. I came across the following formula (Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{...

Way to show $x^n + y^n = z^n$ factorises as $(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n$

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能否分解成除了第一项因式（x-y）之外，其余部分全部由（x+y）、（x*y）和它们的组合所构成的形式？即令a…

影视仓. 片头片尾快捷标记，最新exo支持软硬解，无缝换线路换源