Question

Me ajudem na questão 22,a resposta tem que dar 20 sob 3.

Answer 1

Bom dia Lucas!

Solução!

Para resolver essa questão a principio vamos inverter todo valor que estiver expoente negativo, e em seguida substituir os valores numéricos.

$a=2 \\\\ b=-1$

Vamos determinar o valor de x.

$x=\left ( \dfrac{a^{-2}+b^{-2} }{a^{-1} +b^{-1} } \right )\\\\\\\ x=\left ( \dfrac{ \dfrac{1}{a^{2} } + \dfrac{1}{b^{2} } }{ \dfrac{1}{a} + \dfrac{1}{b} } \right )\\\\\\ x=\left ( \dfrac{ \dfrac{1}{(2)^{2} } + \dfrac{1}{(-1)^{2} } }{ \dfrac{1}{2} + \dfrac{1}{-1} } \right )\\\\\\\ x=\left ( \dfrac{ \dfrac{1}{4 } + \dfrac{1}{1 } }{ \dfrac{1}{2} + \dfrac{1}{-1} } \right )$

$x=\left ( \dfrac{ \dfrac{5}{4 } }{ -\dfrac{1}{2} } \right )\\\\\\\\ x=\left ( -\dfrac{10}{4} \right )$

Vamos determinar y

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

$y=\left ( \dfrac{2}{ \dfrac{1}{4 } + \dfrac{1}{-1 } } \right )\\\\\\\\\\\\\\ y=\left ( \dfrac{2}{ -\dfrac{3}{4 } } \right )\\\\\\\\\\ y=\left ( -\frac{8}{3} \right )$

Sendo 

$x\times y$

$x.y= -\dfrac{10}{4} \times -\dfrac{8}{3}$


Simplificando a fração por 4.


$x.y= -\dfrac{80}{12} \\\\\\\ x.y=\dfrac{20}{3}$


Bons estudos!