Respostas

  • Usuário do Brainly
2013-06-03T19:14:45-03:00

Temos que:

 

\binom{\text{n}}{\text{k}}=\dfrac{\text{n}!}{\text{k}!\cdot(\text{n}-\text{k})!}

 

Analogamente, temos:

 

\binom{\text{x}}{2}=21

 

\dfrac{\text{x}!}{2!\cdot(\text{x}-2)!}=21

 

Observe que:

 

\text{x}!=\text{x}\cdot(\text{x}-1)\cdot(\text{x}-2)!

 

Logo:

 

\dfrac{\text{x}\cdot(\text{x}-1)\cdot(\text{x}-2)!}{2!\cdot(\text{x}-2)!}=21

 

\dfrac{\text{x}\cdot(\text{x}-1)}{2!}=21

 

Donde, obtemos:

 

\text{x}\cdot(\text{x}-1)=42

 

\text{x}^2-\text{x}-42=0

 

\text{x}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-42)}}{2\cdot1}=\dfrac{1\pm13}{2}

 

\text{x}'=\dfrac{1+13}{2}=7

 

\text{x}"=\dfrac{1-13}{2}=-6

 

Desse modo, como \text{x}\in\mathbb{N}, temos que \text{x}=7.

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