\(x\ne2\)
Áp dụng HĐT \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
\(\left(\frac{x-3}{x-2}\right)^3-\left(x-3\right)^3=16\)
\(\Leftrightarrow\left(\frac{\left(x-3\right)-\left(x-3\right)\left(x-2\right)}{x-2}\right)^3+\frac{3\left(x-3\right)^2}{\left(x-2\right)}\left(\frac{x-3}{x-2}-x+3\right)=16\)
\(\Leftrightarrow\left(\frac{\left(x-3\right)\left(3-x\right)}{\left(x-2\right)}\right)^3+\frac{3\left(x-3\right)^2}{x-2}\left(\frac{\left(x-3\right)\left(3-x\right)}{x-2}\right)=16\)
\(\Leftrightarrow\left(-\frac{\left(x-3\right)^2}{x-2}\right)^3-3.\left(\frac{\left(x-3\right)^2}{x-2}\right)^2=16\)
Đặt \(\frac{\left(x-3\right)^2}{x-2}=a\)
\(-a^3-3a^2=16\Leftrightarrow a^3+3a^2+16=0\Rightarrow a=-4\)
\(\Rightarrow\frac{\left(x-3\right)^2}{x-2}=-4\Leftrightarrow x^2-2x+1=0\Rightarrow x=1\)
