Đáp án: $x=12$
Giải thích các bước giải:
ĐKXĐ: $x\ge 2$
Ta có:
$C^0_x+C^{x-1}_x+C^{x-2}_x=79$
$\to 1+\dfrac{x!}{(x-1)!\cdot (x-(x-1))!}+\dfrac{x!}{(x-2)!\cdot (x-(x-2))!}=79$
$\to \dfrac{x!}{(x-1)!\cdot 1!}+\dfrac{x!}{(x-2)!\cdot 2!}=78$
$\to \dfrac{x!}{(x-1)!}+\dfrac{x!}{(x-2)!\cdot 2!}=78$
$\to \dfrac{x!}{(x-2)!}(\dfrac{1}{x-1}+\dfrac{1}{2!})=78$
$\to \dfrac{x!}{(x-2)!}(\dfrac{1}{x-1}+\dfrac{1}{2})=78$
$\to \dfrac{x!}{(x-2)!}\cdot\dfrac{2+x-1}{2(x-1)}=78$
$\to \dfrac{x!}{(x-2)!}\cdot\dfrac{x+1}{2(x-1)}=78$
$\to \dfrac{(x+1)!}{2(x-1)!}=78$
$\to \dfrac{(x+1)!}{2(x-1)!}=78$
$\to \dfrac{(x+1)!}{(x-1)!}=156$
$\to (x+1)x=156$
$\to x=12$
