Question

Tìm số tự nhiên x thỏa

\(A^2_{x-2}+C^{x-2}_x=101\)

Answer 1

ĐK: \(x\ge4\)

\(\dfrac{\left(x-2\right)!}{\left(x-4\right)!}+\dfrac{x!}{\left(x-2\right)!.2!}=101\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)+\dfrac{x\left(x-1\right)}{2}=101\)

\(\Leftrightarrow3x^2-11x-190=0\)

\(\Rightarrow x=10\)