\(\frac{2AB^2}{BC^2}-1=\frac{AB^2+AC^2-BC^2}{BC^2}=\frac{AM^2+BM^2+AM^2+MC^2+2AM.MC-MC^2-BM^2}{BC^2}=\frac{2AM^2+2AM.MC}{BC^2}=\frac{2AM\left(AM+MC\right)}{BC^2}=\frac{2AM.AC}{BC^2}\)
Nếu\(\frac{AM}{AC}=\frac{2AM.AC}{BC^2}\Leftrightarrow\frac{1}{AC}=\frac{2AC}{BC^2}\Leftrightarrow2AC^2=BC^2\) (vô lí)
Xem cách mk làm xem chỗ nào sai dùm nha