検索キーワード「right triangle」に一致する投稿を日付順に表示しています。 関連性の高い順 すべての投稿を表示
検索キーワード「right triangle」に一致する投稿を日付順に表示しています。 関連性の高い順 すべての投稿を表示

上 the identity (x^2 y^2)^2=(x^2-y^2)^2 (2xy)^2 can be used to generate pythagorean triples 250437

X and y are positive integers;Write a(x)/ b(x) in the form q(x) r(x)/ b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division,I am trying to solve the equation $$ (x^2y^2)y' 2xy = 0 $$ I have rearranged to get $$ y' = f(x,y) $$ where $$ f(x,y) = \frac{2xy}{x^2y^2} $$ From here I tried to use a trick that I learned Stack Exchange Network Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn,

Pythagorean Triples

Pythagorean Triples

The identity (x^2 y^2)^2=(x^2-y^2)^2 (2xy)^2 can be used to generate pythagorean triples

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