Question

Determine o valor de A × B

X=3
Y=-2

Equação na imagem.

Answer 1

Eae Jean,

se $A= \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}~~e~~B=\left( \dfrac{x^{-2}+y^{-2}}{x}\right)$

podemos usar algumas das propriedades da exponenciação, e em seguida substituirmos os valores de x e y:

$\boxed{a^{-1}\Rightarrow \dfrac{1}{a^1}\Rightarrow \dfrac{1}{a}}\\ \boxed{\left( \dfrac{a}{b}\right)^{-c}\Rightarrow \dfrac{b}{a}^c}$

$\left( \dfrac{ x^{-2}+y^{-2}}{x^{-1}+y^{-1}} \right)\cdot\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1} =\left( \dfrac{ \dfrac{1}{x^2}+ \dfrac{1}{y^2} }{ \dfrac{1}{x}+ \dfrac{1}{y} }\right)\cdot\left( \dfrac{ \dfrac{1}{x^2}+ \dfrac{1}{y^2} }{x}\right)^{-1}$

$\left( \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\right)\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1}=\left( \dfrac{ \dfrac{1}{x^2}+ \dfrac{1}{y^2} }{ \dfrac{1}{x}+ \dfrac{1}{y} }\right)\cdot\left( \dfrac{x}{ \dfrac{1}{x^2}+ \dfrac{1}{y^2} }\right)^1$

$\left( \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\right)\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1}= \left\{\dfrac{ \left(\dfrac{1}{3^2}\right)+ \dfrac{1}{(-2)^2} }{ \dfrac{1}{3}+ \left(-\dfrac{1}{2}\right) }\right\}\cdot\left\{ \dfrac{3}{ \dfrac{1}{3^2}+ \dfrac{1}{(-2)^2} }\right\}$

$\left( \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\right)\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1}=\left( \dfrac{ \dfrac{1}{9}+ \dfrac{1}{4} }{ \dfrac{1}{3}- \dfrac{1}{2} }\right)\left( \dfrac{3}{ \dfrac{1}{9}+ \dfrac{1}{4}}\right)$

$\left( \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\right)\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1}= \left(\dfrac{ \dfrac{13}{36} }{- \dfrac{1}{6} }\right)\cdot\left( \dfrac{3}{ \dfrac{13}{36} }\right)$

$\left( \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\right)\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1}=\left(- \dfrac{13}{6}\right)\cdot\left( \dfrac{108}{13}\right)\\\\\\ \left( \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\right)\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1}= -\dfrac{1.404}{78}\\\\\\ \Large\boxed{\boxed{\left( \dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\right)\left( \dfrac{x^{-2}+y^{-2}}{x}\right)^{-1}=-18}}$

Tenha ótimos estudos ;D