Đặt \(\frac{8-x}{x-1}\) = y ( y thuộc Z )
Khi đó phương trình trở thành :
xy ( x - y ) = 15
<=> x2y - xy2 = 15
ĐKXĐ:...
\(\Leftrightarrow\frac{8x-x^2}{x-1}\left(\frac{x^2-8}{x-1}\right)=15\)
\(\Leftrightarrow\frac{8x-x^2}{x-1}\left(\frac{x^2-8x+8x-8}{x-1}\right)=15\)
\(\Leftrightarrow-\left(\frac{x^2-8x}{x-1}\right)\left(\frac{x^2-8x}{x-1}+8\right)=15\)
Đặt \(\frac{x^2-8x}{x-1}=a\)
\(-a\left(a+8\right)=15\Leftrightarrow a^2+8a+15=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-3\\a=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x^2-8x}{x-1}=-3\\\frac{x^2-8x}{x-1}=-5\end{matrix}\right.\) \(\Leftrightarrow...\)
