Bài 1:
a) $\dfrac{5^6.20^4}{25^5.4^3}$
$= \dfrac{5^6.(5.4)^4}{(5^2)^5.4^3}$
$= \dfrac{5^6.5^4.4^4}{5^{10}.4^3}$
$= \dfrac{5^{10}.4^4}{5^{10}.4^3}$
$= 4$
b) $3(2x - 3)^2 = 48$
$\Leftrightarrow (2x - 3)^2 = 16$
$\Leftrightarrow |2x - 3| = 4$
$\Leftrightarrow \left[\begin{array}{l}2x - 3 = 4\\2x - 3 = -4\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x= 7\\2x = -1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= \dfrac{7}{2}\\x = -\dfrac{1}{2}\end{array}\right.$
Bài 2:
a) Ta có:
$x//y$
$m\perp x$
$\Rightarrow m\perp y$
$\Rightarrow \widehat{ABD}=90^o$
b) Ta có:
$x//y$
$\Rightarrow \widehat{BDC} + \widehat{C_1} = 180^o$ (trong cùng phía)
$\Rightarrow \widehat{C_1} = 180^o -\widehat{BDC} = 180^o - 57^o = 123^o$
Ta có:
$\widehat{C_2} = \widehat{BDC}$ (so le trong)
$\Rightarrow \widehat{C_2} = 57^o$
$\widehat{C_3} = \widehat{BDC}$ (đồng vị)
$\Rightarrow \widehat{C_3} = 57^o$
