Question

please provide me the short trick of this question​

Answer 1

Answer :-

(d) (x + 4)/x

Solution :-

$\sf \bigg(1 \dfrac{1}{x} \bigg) \bigg(1 + \dfrac{1}{x + 1} \bigg) \bigg(1 + \dfrac{1}{x + 2} \bigg) \bigg(1 + \dfrac{1}{x + 3} \bigg)$

Taking LCM

$\sf = \bigg( \dfrac{x + 1}{x} \bigg) \bigg( \dfrac{1(x + 1) + 1}{x + 1} \bigg) \bigg( \dfrac{1(x + 2) + 1}{x + 2} \bigg) \bigg( \dfrac{1(x + 3) + 1}{x + 3} \bigg)$

$\\ \sf = \bigg( \dfrac{x + 1}{x} \bigg) \bigg( \dfrac{x + 1 + 1}{x + 1} \bigg) \bigg( \dfrac{x + 2 + 1}{x + 2} \bigg) \bigg( \dfrac{x + 3 + 1}{x + 3} \bigg)$

$\\ \sf = \bigg( \dfrac{ \cancel{x + 1}}{x} \bigg) \bigg( \dfrac{ \cancel{x +2}}{ \cancel{x + 1}} \bigg) \bigg( \dfrac{ \cancel{x + 3}}{ \cancel{x + 2}} \bigg) \bigg( \dfrac{x + 4}{ \cancel{x + 3}} \bigg)$

$\\ \sf = \dfrac{x + 4}{x}$

Therefore (d) (x + 4)/x is the answer.

Answer 2

Answer

$\huge{ \sf{1. \frac{1}{x} }. (1 + \frac{1}{x + 1}) .(1 + \frac{1}{x + 2}).(1 + \frac{1}{x + 3}) } \\ \\ \implies \: \sf{ \frac{x+1}{x}. \frac{(x + 1) + 1}{x + 1}. \frac{(x + 2) + 1}{x + 2}. \frac{(x + 3) + 1}{x + 3} } \\ \\ \implies \: \sf{ \frac{\cancel{x+1}}{x}. \frac{ \cancel{x + 2}}{\cancel{x + 1}}. \frac{ \cancel{x + 3}}{ \cancel{x + 2}}. \frac{x + 4}{ \cancel{x + 3}} } \\ \\ \implies \: \underline{\boxed{\sf{ \frac{x + 4}{x}}}}$

Thus,the correct option is (d)