Bài 5: Bảng căn bậc hai

\(CH=\dfrac{AH^2}{BH}=16\left(cm\right)\)

\(AB=\sqrt{BH\cdot BC}=\sqrt{9\cdot25}=15\left(cm\right)\)

AC=20(cm)

\(\widehat{B}\simeq37^0\)

\(\widehat{C}\simeq53^0\)

Áp dụng HTL:

\(CH=\dfrac{AH^2}{BH}=16\left(cm\right)\Rightarrow BC=BH+BC=25\left(cm\right)\)

\(\Rightarrow\left\{{}\begin{matrix}AB=\sqrt{BH\cdot BC}=15\left(cm\right)\\AC=\sqrt{CH\cdot BC}=20\left(cm\right)\end{matrix}\right.\)

\(\sin B=\dfrac{AC}{BC}=\dfrac{20}{25}=\dfrac{4}{5}\approx53^0\Rightarrow\widehat{B}\approx53^0\\ \widehat{C}=90^0-\widehat{B}\approx90^0-53^0=37^0\)

\(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{7\left(4+\sqrt{2}\right)}{14}+\dfrac{\sqrt{3}}{3}\\ =\sqrt{6}-\dfrac{4+\sqrt{2}}{2}+\dfrac{\sqrt{3}}{3}=\dfrac{6\sqrt{6}-12-3\sqrt{2}+2\sqrt{3}}{6}\)

\(=\left[\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right]\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\left(a,b>0;a\ne b\right)\\ =\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)

\(=\sqrt{5+2\sqrt{3}\cdot\sqrt{5}+3}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

Bài 2:

\(AB^2=BH\cdot BC\)

\(\Leftrightarrow BH\left(BH+3.2\right)=9\)

hay BH=1,8(cm)
\(\Leftrightarrow AH=2,4\left(cm\right)\)

hay AC=4cm