If $$\dfrac{1}{6!}+ \dfrac{1}{7!}= \dfrac{x}{8!}$$ find x.

Asked by Abhisek | 2 years ago |  88

##### Solution :-

$$\dfrac{1}{6!}+ \dfrac{1}{7!}= \dfrac{x}{8!}$$

$$\dfrac{1}{6!}+ \dfrac{1}{7!\times 6!}= \dfrac{x}{8\times 7\times 6!}$$

$$\dfrac{1}{6!}+ (1+\dfrac{1}{7!})= \dfrac{x}{8\times 7\times 6!}$$

$$1+\dfrac{1}{7}= \dfrac{x}{8\times 7}$$

$$\dfrac{8}{7}= \dfrac{x}{8\times 7}$$

$$\dfrac{8\times 8\times 7}{7}$$

= 64

Answered by Pragya Singh | 2 years ago

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