We have, $\frac{{({x^2} + 5x + 6)({x^2} + 9x + 20)}}{{({x^2} – 5x + 6)({x^2} – 9x + 20)}} = 1$

$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} – 5x + 6}} = \frac{{{x^2} – 9x + 20}}{{{x^2} + 5x + 6}} = \frac{{({x^2} + 9x + 20) – ({x^2} – 9x + 20)}}{{({x^2} – 5x + 6) – ({x^2} + 5x + 6)}} = \frac{{18x}}{{ – 10x}} = – \frac{{9}}{{5}}$
(assuming $x \ne 0$)

$ \Rightarrow \frac{{{x^2} + 9x + 20}}{{{x^2} – 5x + 6}} – 1 = – \frac{9}{5} – 1$

$ \Rightarrow \frac{{14x + 14}}{{{x^2} – 5x + 6}} = – \frac{{14}}{5}$

$ \Rightarrow \frac{{x + 1}}{{{x^2} – 5x + 6}} = – \frac{1}{5}$

$ \Rightarrow {x^2} – 5x + 6 + 5x + 5 = 0$ Or ${x^2} + 11 = 0$

So, no real solution if $x \neq 0$.

If x = 0, LHS of the original equation = $\frac{{2 \times 3 \times 4 \times 5}}{{( – 2) \times ( – 3) \times ( – 4) \times ( – 5)}}$ = 1 = RHS

Thus, x = 0 is the only real solution.

Select correct option(s).

(A) a, b, c are in A.P.
(B) a, b, c are in G.P.
(C) a, b, c are in H.P.
(D) a+b+c = bc

Solution

We have, $a + \frac{1}{{b + \frac{1}{c}}} = \frac{{115}}{{16}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{115}}{{15 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{115}}{{5 \times 3 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{16 \times 7 + 3}}{{5 \times 3 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = 7 + \frac{3}{{5 \times 3 + 1}}$

$\therefore \{ a,b,c\} \equiv \{ 7,5,3\} $

Hence, (A) & (D).

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.

Solve for $x \in R$, $x^9 +x^6 = 36$

Solution

Let, $x^3 =t$

We have, $t^3 +t^2 = 36$

$\therefore t^3 +t^2 -36 = 0$ Continue reading $x^9 +x^6 = 36$ →