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### Solve for $x \in \mathbb{R}$, $\frac{{(x + 2)(x + 3)(x + 4)(x + 5)}}{{(x - 2)(x - 3)(x - 4)(x - 5)}} = 1$

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.

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$