Question

Q in photo please answer it very difficult qno 28

Answer 1

Solution :-

It is provided :-

$a + \dfrac{1}{a} + 2 = 0$

By further solving our equation :-

$\implies \dfrac{ a^2 + 1 + 2a }{a} = 0$

$\implies a^2 + 2a + 1 = 0$

$\implies a^2 + a + a + 1 = 0$

$\implies a(a + 1) + 1(a + 1) = 0$

$\implies (a+1)(a+1) = 0$

$\implies a = -1 , -1$

So the value of a = -1

Now the value of

$a^{37} - \dfrac{1}{a^{100}}$

$= (-1)^{37} - \dfrac{1}{(-1)^{100}}$

$=( -1) - \dfrac{1}{1}$

$=( -1 ) - 1$

$= -2$

Answer 2

Answer:

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