Ta có:      BD=AB.sinA=a.sin(alpha)   

               AD=AB.cosA=a.cos(alpha)

=>S=2SABD

      =BD.AD=a².sin(alpha).cos(alpha)

tự làm đi

Vẽ đường cao CH. Ta có:

\(\hept{\begin{cases}BH+AH=14\\BH^2+CH^2=225\\AH^2+CH^2=169\end{cases}\Rightarrow\hept{\begin{cases}BH+AH=14\\BH^2-AH^2=56\end{cases}\Leftrightarrow}\hept{\begin{cases}BH+AH=14\\\left(BH+AH\right)\left(BH-AH\right)=56\end{cases}\Leftrightarrow}\hept{\begin{cases}BH+AH=14\\BH-AH=4\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}BH=9\\AH=5\end{cases}\Rightarrow\hept{\begin{cases}\widehat{A}=cos^{-1}\frac{5}{13}\approx67^023'\\\widehat{B}=cos^{-1}\frac{9}{15}\approx53^08'\\\widehat{C}\approx180^0-\left(67^023'+53^08'\right)=59^029'\end{cases}}}\)

\(\sqrt[3]{x-9}=\left(x-3\right)^3+6\)

\(\Leftrightarrow\sqrt[3]{x-9}+2=\left(x-3\right)^3+8\)

\(\Leftrightarrow\frac{x-9+8}{\sqrt[3]{x-9}^2-2\sqrt[3]{x-9}+8}=x^3-9x^2+27x-19\)

\(\Leftrightarrow\frac{x-1}{\sqrt[3]{x-9}^2-2\sqrt[3]{x-9}+8}-\left(x-1\right)\left(x^2-8x+19\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt[3]{x-9}^2-2\sqrt[3]{x-9}+8}-\left(x^2-8x+19\right)\right)=0\)

Suy ra x=1

ĐKXĐ: \(-1\le x\le8\)

Đặt \(a=\sqrt{1+x};b=\sqrt{8-x}\). Ta có: \(\hept{\begin{cases}a^2+b^2=9\\a+b+ab=3\end{cases}\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2-2ab=9\\a+b+ab=3\end{cases}\Leftrightarrow}\hept{\begin{cases}ab=0\\a+b=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}a=0;b=3\\a=3;b=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{1+x}=0;\sqrt{8-x}=3\\\sqrt{1+x}=3;\sqrt{8-x}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=8\end{cases}}}\)