PT<=> [tex][\frac{x-3}{x-2}-(x-3)]^3+3.\frac{x-3}{x-2}.(x-3)[\frac{x-3}{x-2}-(x-3)]=16\Leftrightarrow [\frac{-(x-3)^2}{x-2}]^3+3.\frac{(x-3)^2}{x-2}.\frac{-(x-3)^2}{x-2}=16 \Leftrightarrow [\frac{(x-3)^2}{x-2}]^3+3.[\frac{(x-3)^2}{x-2}]^2+16=0[/tex]
Đặt t=(x-3)^2/x-2
Pt-->t^3+3t^2+16=0
Tự giải tiếp