Question

giải hệ phương trình sau \(\left\{{}\begin{matrix}C^{y+1}_{x+1}=C_{x+1}^y\\3C^{y+1}_{x+1}=5C^{y-1}_{x+1}\end{matrix}\right.\)

Answer 1

ĐK: \(x\ge y\ge1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{\left(x+1\right)!}{\left(y+1\right)!\left(x-y\right)!}=\frac{\left(x+1\right)!}{y!\left(x-y+1\right)!}\\\frac{3\left(x+1\right)!}{\left(y+1\right)!\left(x-y\right)!}=\frac{5\left(x+1\right)!}{\left(y-1\right)!\left(x-y+2\right)!}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+1=x-y+1\\3\left(x-y+2\right)\left(x-y+1\right)=5y\left(y+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\3\left(x-y+2\right)\left(x-y+1\right)=5y\left(y+1\right)\end{matrix}\right.\)

\(\Rightarrow3\left(y+2\right)\left(y+1\right)=5y\left(y+1\right)\)

\(\Leftrightarrow3y+6=5y\Rightarrow y=3\Rightarrow x=6\)