If x∈N and (x+2)!(2x$$−$$1)!⋅(2x+1)!(x+3)!$$=727$$,then x is equal to

(2x+1)(2x)x+3=727$$⇒$$72x2+x=36(x+3)⇒$$14x2$$−$$29x$$−$$108=0$$⇒$$(14x+27)(x$$−$$4)=0$$⇒$$x=4$$

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Question

If $x \in N$ and $\frac{(x + 2)!}{(2 x - 1)!} \cdot \frac{(2 x + 1)!}{(x + 3)!} = \frac{72}{7},$ then $x$ is equal to

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Solution

$\frac{(2 x + 1) (2 x)}{x + 3} = \frac{72}{7}$
$\begin{array}{l} \Rightarrow 7 (2 x^{2} + x) = 36 (x + 3) \\ \Rightarrow 14 x^{2} - 29 x - 108 = 0 \\ \Rightarrow (14 x + 27) (x - 4) = 0 \Rightarrow x = 4 \end{array}$

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