Bài 4c:
Ta có:
$sinC = \dfrac{AB}{BC}$
$sinC = sin\widehat{BHE} = \dfrac{BE}{BH}$
$sinC = sin\widehat{BAH} = \dfrac{BH}{AB}$
$\Rightarrow sin^3C = \dfrac{AB}{BC}\cdot \dfrac{BE}{BH}\cdot \dfrac{BH}{AB} = \dfrac{BE}{BC}$
$\Rightarrow BE = BC.sin^3C$
Bài 5:
Sửa đề:
$S = \dfrac{1}{2\sqrt{1} + 1\sqrt{2}} + \dfrac{1}{3\sqrt{2} + 2\sqrt{3}} + \cdots +\dfrac{1}{4080400\sqrt{4080399} + 4080399\sqrt{4080400}}$
Ta có:
$\star \, \dfrac{1}{2\sqrt{1} + 1\sqrt{2}}=\dfrac{1}{\sqrt{2}.\sqrt{1}(\sqrt{2} + \sqrt{1})} = \dfrac{\sqrt{2} - \sqrt{1}}{\sqrt{2}.\sqrt{1}}=\dfrac{\sqrt{2}}{\sqrt{2}.\sqrt{1}}-\dfrac{\sqrt{1}}{\sqrt{2}.\sqrt{1}}=\dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{2}}$
$\star \, \dfrac{1}{3\sqrt{2} + 2\sqrt{3}}=\dfrac{1}{\sqrt{3}.\sqrt{2}(\sqrt{3} + \sqrt{2})} = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3}.\sqrt{2}}=\dfrac{\sqrt{3}}{\sqrt{3}.\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}.\sqrt{2}}=\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{3}}$
$\cdots$
$\star \, \dfrac{1}{4080400\sqrt{4080399} + 4080399\sqrt{4080400}}=\dfrac{1}{\sqrt{4080400}.\sqrt{4080399}(\sqrt{4080400} + \sqrt{4080399})} = \dfrac{\sqrt{4080400} - \sqrt{4080399}}{\sqrt{4080400}.\sqrt{4080399}}=\dfrac{\sqrt{4080400}}{\sqrt{4080400}.\sqrt{4080399}}-\dfrac{\sqrt{4080399}}{\sqrt{4080400}.\sqrt{4080399}}=\dfrac{1}{\sqrt{4080399}} - \dfrac{1}{\sqrt{4080400}}$
$\Rightarrow S = \dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{3}} + \cdots + \dfrac{1}{\sqrt{4080399}} - \dfrac{1}{\sqrt{4080400}}$
$= 1 - \dfrac{1}{\sqrt{4080400}} = 1 - \dfrac{1}{2020}$
$= \dfrac{2019}{2020}$
