Đáp án:

B4:

 a) $A=0$

b) $B = \dfrac{{{{21.6}^6}}}{{1 + {6^8}}}$

B2: 

$x =  - 1;y =  - 2;z = 3$

Giải thích các bước giải:

$\begin{array}{l}
a)A = \left( {\dfrac{{ - 5}}{7} + \dfrac{8}{{11}}} \right):\dfrac{7}{{11}} + \left( {\dfrac{{ - 2}}{7} + \dfrac{3}{{11}}} \right):\dfrac{7}{{11}}\\
 = \left( {\dfrac{{ - 5}}{7} + \dfrac{8}{{11}} + \dfrac{{ - 2}}{7} + \dfrac{3}{{11}}} \right).\dfrac{{11}}{7}\\
 = \left( {\left( {\dfrac{{ - 5}}{7} + \dfrac{{ - 2}}{7}} \right) + \left( {\dfrac{8}{{11}} + \dfrac{3}{{11}}} \right)} \right).\dfrac{{11}}{7}\\
 = \left( { - 1 + 1} \right).\dfrac{{11}}{7}\\
 = 0\\
b)B = \dfrac{{{2^{19}}{{.27}^3} + {{15.4}^9}{{.9}^4}}}{{{6^2}{{.2}^{10}} + {{12}^{10}}}}\\
 = \dfrac{{{2^{19}}{{.3}^9} + {{3.5.2}^{18}}{{.3}^8}}}{{{2^2}{{.3}^2}{{.2}^{10}} + {2^{20}}{{.3}^{10}}}}\\
 = \dfrac{{{2^{19}}{{.3}^9} + {{5.2}^{18}}{{.3}^9}}}{{{2^{12}}{{.3}^2} + {2^{20}}{{.3}^{10}}}}\\
 = \dfrac{{{2^{18}}{{.3}^9}\left( {2 + 5} \right)}}{{{2^{12}}{{.3}^2}\left( {1 + {2^8}{{.3}^8}} \right)}}\\
 = \dfrac{{{2^6}{{.3}^7}.7}}{{1 + {6^8}}}\\
 = \dfrac{{{{21.6}^6}}}{{1 + {6^8}}}\\
2){\left( {x + 1} \right)^{30}} + {\left( {y + 2} \right)^4} + {\left( {z - 3} \right)^{2020}} = 0\\
\left\{ \begin{array}{l}
{\left( {x + 1} \right)^{30}} = {\left( {{{\left( {x + 1} \right)}^{15}}} \right)^2} \ge 0,\dforall x\\
{\left( {y + 2} \right)^4} = {\left( {{{\left( {y + 2} \right)}^2}} \right)^2} \ge 0,\dforall y\\
{\left( {z - 3} \right)^{2020}} = {\left( {{{\left( {z - 3} \right)}^{1010}}} \right)^2} \ge 0,\dforall z
\end{array} \right.\\
 \Rightarrow {\left( {x + 1} \right)^{30}} = {\left( {y + 2} \right)^4} = {\left( {z - 3} \right)^{2020}} = 0\\
 \Leftrightarrow x + 1 = y + 2 = z - 3 = 0\\
 \Leftrightarrow x =  - 1;y =  - 2;z = 3
\end{array}$

Vậy $x =  - 1;y =  - 2;z = 3$