Question

Alguém pode me explicar essa questão por favor.​

Answer 1

Resposta:

já há pergunta igual!

https://brainly.com.br/tarefa/1657188

Explicação passo-a-passo:

Answer 2

Resposta:

Item d)

Explicação passo-a-passo:

Veja que

$a^{-1}=\dfrac{1}{a}\\\\b^{-1}=\dfrac{1}{b}\\\\(a^{-1}+b^{-1})^{-2}=(\dfrac{1}{a^{-1}+b^{-1}})^{2}\\\\\\(a^{-1}+b^{-1})^{-2}=\dfrac{1^{2}}{(a^{-1}+b^{-1})^{2}}\\\\\\(a^{-1}+b^{-1})^{-2}=\dfrac{1}{(\frac{1}{a}+\frac{1}{b})^{2}}\\\\\\(a^{-1}+b^{-1})^{-2}=\dfrac{1}{(\frac{1}{a})^{2}+2.\frac{1}{a}.\frac{1}{b}+(\frac{1}{b})^{2}}\\\\\\(a^{-1}+b^{-1})^{-2}=\dfrac{1}{\frac{1}{a^{2}}+\frac{2}{a.b}+\frac{1}{b^{2}}}\\\\\\mmc(a^{2},ab,b^{2})=a^{2}b^{2}$

$(a^{-1}+b^{-1})^{-2}=\dfrac{1}{\frac{b^{2}+2ab+a^{2}}{a^{2}b^{2}}}\\\\\\(a^{-1}+b^{-1})^{-2}=\dfrac{a^{2}b^{2}}{{b^{2}+2ab+a^{2}}}\\\\\\(a^{-1}+b^{-1})^{-2}=\dfrac{a^{2}b^{2}}{(a+b)^{2}}}$