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$a + b + c = 2022$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{2022}}$

$S = \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = ?$

We have, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{{2022}} = \frac{1}{{a + b + c}}$

$ \Rightarrow \frac{1}{a} + \frac{1}{b} + \left( {\frac{1}{c} - \frac{1}{{a + b + c}}} \right) = 0$

$ \Rightarrow \frac{{a + b}}{{ab}} + \frac{{a + b}}{{(a + b + c)c}} = 0$

$ \Rightarrow (a + b)\frac{{(ab + bc + ca + {c^2})}}{{abc(a + b + c)}} = 0$

$ \Rightarrow (a + b)\frac{{(b + c)(c + a)}}{{abc(a + b + c)}} = 0$

$ \Rightarrow (a + b)(b + c)(c + a) = 0$

$a + b = 0$ Or $b + c = 0$ Or $c + a = 0$

If $a + b = 0$, $a + b + c = 0 + c = c = 2022$

$S = \frac{1}{{{a^{2023}}}} + \frac{1}{{{b^{2023}}}} + \frac{1}{{{c^{2023}}}} = \frac{1}{{{a^{2023}}}} + \frac{1}{{{{( - a)}^{2023}}}} + \frac{1}{{{{2022}^{2023}}}} = \frac{1}{{{{2022}^{2023}}}}$

Due to symmetry, other cases would yield the same final answer.

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$