Bài 1:
a. \(\sqrt{\frac{25}{144}}=\sqrt{(\frac{5}{12})^2}=\frac{5}{12}\)
b. \(\sqrt{2\frac{7}{81}}=\sqrt{\frac{169}{81}}=\sqrt{(\frac{13}{9})^2}=\frac{13}{9}\)
c. \(\frac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\frac{12,5}{0,5}}=\sqrt{25}=\sqrt{5^2}=5\)
d. \(\frac{\sqrt{192}}{\sqrt{12}}=\sqrt{\frac{192}{12}}=\sqrt{16}=\sqrt{4^2}=4\)
e. \(\sqrt{\frac{-49}{-81}}=\sqrt{\frac{49}{81}}=\sqrt{(\frac{7}{9})^2}=\frac{7}{9}\)
f.
\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\frac{16a^4b^6}{128a^6b^6}}=\sqrt{\frac{1}{8a^2}}=\sqrt{2(\frac{1}{4a})^2}=\frac{1}{4a}\sqrt{2}\)
Bài 2:
Ta có: $\sin C=\frac{AB}{BC}$
$\Rightarrow BC=\frac{AB}{\sin C}=\frac{21}{\sin 40^0}=32,7$ (cm)
$\tan C=\frac{AB}{AC}\Rightarrow AC=\frac{AB}{\tan C}=\frac{21}{\tan 40^0}=25,0$ (cm)
$\widehat{ABD}=\frac{1}{2}\widehat{B}=\frac{1}{2}(90^0-\widehat{C})=\frac{1}{2}(90^0-40^0)=25^0$
$\cos 25^0=\cos \widehat{ABD}=\frac{AB}{BD}$
$\Rightarrow BD=\frac{AB}{\cos 25^0}=\frac{21}{\cos 25^0}=23,2$ (cm)
