Sabendo que $\phantom{X}a\,+\,\dfrac{1}{a}\,=\,3\phantom{X}$, calcular o valor de $\phantom{X}a^2\,+\,\dfrac{1}{a^2}\phantom{X}$
resposta:
Resolução:
$\,a\,+\,\dfrac{1}{a}\,=\,3\;\Rightarrow$ $\;\left(a\,+\,\dfrac{1}{a}\right)^2\,=\,9\;\Leftrightarrow$ $\;a^2\,+\,2\,\centerdot\,a\,\centerdot\,\dfrac{1}{a}\,+\,\dfrac{1}{a^2}\,=\,9\;\Leftrightarrow$ $\;a^2\,+\,2\,+\,\dfrac{1}{a^2}\,=\,9\;\Leftrightarrow$ $\;a^2\,+\,\dfrac{1}{a^2}\,=\,9\,-\,2\,=$ $\,7$
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