\(AB^2=BH.BC\) (theo hệ thức lượng trong tam giác vuông)

\(\Rightarrow BC=\dfrac{AB^2}{BH}=\dfrac{100^2}{5}=2000\left(cm\right)\)

\(\Rightarrow HC=BC-HB=2000-5=1995\left(cm\right)\)

\(AH^2=BH.HC\Leftrightarrow AH^2=1995.5\Leftrightarrow AH=5\sqrt{399}\)

B

\(a)P=\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt[]{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{\sqrt{4.8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{\sqrt{4}\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}\)

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

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

\(b)Q=\dfrac{3-2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

\(=\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

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

\(=-2\)

3. getting up early and preparing breakfast in the cold winter

4. watching historical films

5. doing motor racing

6. in swimming and sunbathing

7. one hour traveling to his hometown

8. doing any sports or exercises,...

9. making crafts,...

10. going fishing in the rural areas

I. 1. spending

2. listening

3. visiting

4. getting

5. making-eating

6. getting

7. working

8. driving

9. reading

Cho \(a,b\) >0 và  \(a+b\le2\) . Tìm giá trị nhỏ nhất của biểu thức: \(P=\sqrt[]{a\left(b+1\right)}+\sqrt[]{b\left(a+1\right)}\)