Đáp án:
$\begin{array}{l}
a){x^4} + {\left( {y - \dfrac{4}{7}} \right)^{20}} = 0\\
Do:\left\{ \begin{array}{l}
{x^4} = {\left( {{x^2}} \right)^2} \ge 0\\
{\left( {y - \dfrac{4}{7}} \right)^{20}} = {\left\{ {{{\left( {y - \dfrac{4}{7}} \right)}^{10}}} \right\}^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^4} = 0\\
{\left( {y - \dfrac{4}{7}} \right)^{20}} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = \dfrac{4}{7}
\end{array} \right.\\
Vậy\,x = 0;y = \dfrac{4}{7}\\
b){\left( {\dfrac{1}{3}x - 4} \right)^{10}} + {\left( {{y^2} - \dfrac{1}{9}} \right)^{20}} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{3}x - 4 = 0\\
{y^2} - \dfrac{1}{9} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{3}x = 4\\
{y^2} = \dfrac{1}{9}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 12\\
y = \dfrac{1}{3};y = - \dfrac{1}{3}
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left[ {\left( {12;\dfrac{1}{3}} \right);\left( {12; - \dfrac{1}{3}} \right)} \right]\\
c){\left( {x - 5} \right)^{x + 1}} - {\left( {x - 5} \right)^{x + 21}} = 0\\
\Leftrightarrow {\left( {x - 5} \right)^{x + 1}}.\left( {1 - {{\left( {x - 5} \right)}^{20}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( {x - 5} \right)^{x + 1}} = 0\\
{\left( {x - 5} \right)^{20}} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x - 5 = 1\\
x - 5 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 6\\
x = 4
\end{array} \right.\\
Vậy\,x = 4;x = 5;x = 6
\end{array}$
