If \(a - \frac{1}{a} = 1,\) then the value of \(\frac{{{a^2} - a - 1}}{{{a^2} + a + 1}}\) is (a ≠ 0)
If \(a - \frac{1}{a} = 1,\) then the value of \(\frac{{{a^2} - a - 1}}{{{a^2} + a + 1}}\) is (a ≠ 0)
1). 1
2). -1
3). 0
4). 2
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# $(a - \frac{1}{a} = 1)$
⇒ a2 – 1 = a
⇒ a2 – 1 – a = 0
$(\therefore \frac{{{a^2} - a - 1}}{{{a^2} + a + 1}} = 0)$
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