Предмет: Алгебра, автор: kiryanov123345

помогите пж !!!!!!!!!!​

Приложения:

Ответы

Автор ответа: belovaksenia2007
0
((-4х • -4х) - (8у•8у) +12 ху - 2хху +2хуу)/(6х✔️ху + 6у✔️ху +ху✔️ху + ✔️хууу)
Автор ответа: NNNLLL54
0

x>0\ ,\ y>0\\\\\Big(\dfrac{x-y}{x^{3/4}+x^{1/2}y^{1/4}}-\dfrac{x^{1/2}-y^{1/2}}{x^{1/4}+y^{1/4}}\Big)\cdot \Big(\dfrac{y}{x}\Big)^{-1/2}=\\\\\\=\Big(\dfrac{(x^{1/4}-y^{1/4})(x^{1/4}+y^{1/4})(x^{1/2}+y^{1/2})}{x^{1/2}\cdot (x^{1/4}+y^{1/4})}-\dfrac{(x^{1/4}-y^{1/4})(x^{1/4}+y^{1/4})}{x^{1/4}+y^{1/4}}\Big)\cdot \Big(\dfrac{x}{y}\Big)^{1/2}=

=\Big(\dfrac{(x^{1/4}-y^{1/4})(x^{1/2}+y^{1/2})}{x^{1/2}}-(x^{1/4}-y^{1/4})\Big)\cdot \dfrac{x^{1/2}}{y^{1/2}}=\\\\\\=\dfrac{(x^{1/4}-y^{1/4})(x^{1/2}+y^{1/2}-x^{1/2})}{x^{1/2}}\cdot \dfrac{x^{1/2}}{y^{1/2}}=\dfrac{(x^{1/4}-y^{1/4})\cdot y^{1/2}\cdot x^{1/2}}{x^{1/2}\cdot y^{1/2}}=\\\\\\=x^{1/4}-y^{1/4}=\sqrt[4]{x}-\sqrt[4]{y}

Похожие вопросы