Bài 14:
\(1,ĐK:3x+12\ge0\Leftrightarrow x\ge-4\\ 2,=24\sqrt{2x}+6\sqrt{2x}=30\sqrt{2x}\)
Bài 15:
\(1,BC=\sqrt{AB^2+AC^2}=17\left(cm\right)\left(pytago\right)\)
Áp dụng HTL: \(BH=\dfrac{AB^2}{BC}=\dfrac{64}{17}\left(cm\right)\)
\(2,\sin\widehat{ACB}=\dfrac{AB}{BC}=\dfrac{8}{17}\)


