theo định lý pi-ta-go ta có : 

\(BC=\sqrt{12^2+16^2}=20cm\)

áp dụng hệ thức lượng ta có 

\(AB^2=BH.BC\)

=> \(BH=\frac{AB^2}{BC}=\frac{12^2}{20}=\frac{144}{20}\)= 7.2 cm

=>\(CH=BC-BH=20-7,2=12,8cm\)

1+1=2 :)))))

Bằng 2 nhé bạn 

\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3}+1}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}+1}\)

\(=\frac{\left(2\sqrt{2}+\sqrt{6}\right)\left(3-\sqrt{3}\right)+\left(2\sqrt{2}-\sqrt{6}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)

\(=\frac{6\sqrt{2}}{6}=\sqrt{2}\)

\(\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{c^2}{a-1}\ge\frac{\left(a+b+c\right)^2}{a+b+c-3}\)

Ta chứng minh:

\(\frac{\left(a+b+c\right)^2}{a+b+c-3}\ge12\)

\(\Leftrightarrow\frac{\left(a+b+c-6\right)^2}{a+b+c-3}\ge0\left(đúng\right)\)

Vậy có điều phải chứng minh là đúng