Giải thích các bước giải:

 B4:

$\begin{array}{l}
a)\left( {3x - 5} \right)\left( {5 - 3x} \right) + 9{\left( {x + 1} \right)^2} = 30\\
 \Leftrightarrow  - 9{x^2} + 30x - 25 + 9\left( {{x^2} + 2x + 1} \right) = 30\\
 \Leftrightarrow  - 9{x^2} + 30x - 25 + 9{x^2} + 18x + 9 = 30\\
 \Leftrightarrow 48x = 46\\
 \Leftrightarrow x = \frac{{23}}{{24}}\\
b){\left( {x + 4} \right)^2} - \left( {x + 1} \right)\left( {x - 1} \right) = 16\\
 \Leftrightarrow {x^2} + 8x + 16 - \left( {{x^2} - 1} \right) = 16\\
 \Leftrightarrow 8x =  - 1\\
 \Leftrightarrow x = \frac{{ - 1}}{8}
\end{array}$

B5:

$\begin{array}{l}
a)B = {3^{32}} - 1\\
 = {\left( {{3^{16}}} \right)^2} - 1\\
 = \left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\\
 = \left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
 = \left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
 = \left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
 = \left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
 = 2\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\\
 \Rightarrow B > A
\end{array}$

$\begin{array}{l}
b)A = 2011.2013 = \left( {2012 - 1} \right)\left( {2012 + 1} \right) = {2012^2} - 1\\
 \Rightarrow A < B
\end{array}$

B6:

a) Ta có:

$ABCD$ là hình thang cân nên:

$ \Rightarrow \left\{ \begin{array}{l}
\widehat A + \widehat D = {180^0}\\
\widehat A = \widehat B\\
\widehat C = \widehat D
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\widehat A = \widehat B = {110^0}\\
\widehat C = \widehat D = {70^0}
\end{array} \right.$

Vậy ${\widehat A = \widehat B = {{110}^0};\widehat C = \widehat D = {{70}^0}}$

b) Ta có:;

$ABCD$ là hình thang cân nên $ \Rightarrow \widehat D = \widehat C;AD = BC$

Khi đó:

Xét $\Delta AHD;\Delta BKC$ ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AHD} = \widehat {BKC} = {90^0}\\
AD = BC\\
\widehat {ADH} = \widehat {BCK}
\end{array} \right.\\
 \Rightarrow \Delta AHD = \Delta BKC\left( {ch - gn} \right)\\
 \Rightarrow DH = CK
\end{array}$

Vậy ta có đpcm