Question

Tìm x,y\(\in\)N* :  xy²+2xy+x=32y

Answer 1

pt <=> \(x\left(y^2+2y+1\right)=32y\)

\(\Leftrightarrow x\left(y+1\right)^2=32y\)

\(\Leftrightarrow\frac{x}{y}.\left(y+1\right)^2=32\)

do x,y \(\in\)N* => y+1>1

\(\Leftrightarrow\frac{x}{y}.\left(y+1\right)^2=2.4^2=8.2^2\)

TH1: \(\hept{\begin{cases}\frac{x}{y}=2\\y+1=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=3\end{cases}}\)

TH2: \(\hept{\begin{cases}\frac{x}{y}=8\\y+1=2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=8\\y=1\end{cases}}\)

Vậy (x,y)=...