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