Question

\(\text{Giải pt}\)

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

Answer 1

Giải : 

\(A^2_{x-2}+C^{x-2}_x=101\)\(\left(ĐK:\hept{\begin{cases}x\in Z\\x\ge4\end{cases}}\right)\)

\(\Leftrightarrow\frac{\left(x-2\right)!}{\left(x-4\right)!}+\frac{x!}{\left(x-2\right)!2!}=101\)

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

\(\Leftrightarrow2.\left(x-2\right).\left(x-3\right)+x.\left(x-1\right)=202\)

\(\Leftrightarrow2x^2-6x-4x+12+x^2-x-202=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=10\left(tm\right)\\x=\frac{-19}{3}\left(l\right)\end{cases}}\)