`1`
$a)\dfrac{5^6.20^4}{25^5.4^3}$
$=\dfrac{5^6.5^4.4^4}{5^{10}.4^3}$
$=\dfrac{5^{10}.4^4}{5^{10}.4^3}$
$=4$
`b` , $3.(2x-3)^2=48$
$⇔(2x-3)^2=16$
$⇔2x-3=±4$
$⇔$ \(\left[ \begin{array}{l}2x-3=4\\2x-3=-4\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}2x=7\\2x=-1\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x=\dfrac{2}{7}\\x=-\dfrac{1}{2}\end{array} \right.\)
`2`
`a` , Ta có `a` // `b`
`→` `\hat{A}` `=` `\hat{B}`
Mà `\hat{A}` `=` `90^o` `→` `\hat{B}` `=` `90^o`
`b` , Ta có `a` // `b`
→ `\hat{BDC}` `=` `\hat{C3}`
Mà `\hat{BDC}` `=` `57^o`
`→` `\hat{C3}` `=` `57^o`
Có `\hat{C1}` `+` `\hat{C3}` `=` `180^o` (kề bù)
→ `\hat{C1}` `=` `180^o` `-` `57^o`
→ `\hat{C1}` `=` `123^o`
Có `\hat{C1}` `+` `\hat{C2}` `=` `180^o` (kề bù)
→ `\hat{C2}` `=` `180^o` `-` `123^o`
→ `\hat{C2}` `=` `57^o`
