Đáp án:
\[x = \dfrac{5}{3};\,\,\,y = - \dfrac{2}{{15}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{{{\left( {3x - 5} \right)}^{2018}}}}{9} + \dfrac{{{{\left( {3y + 0,4} \right)}^{2020}}}}{3} = 0\\
\Leftrightarrow 9.\left[ {\dfrac{{{{\left( {3x - 5} \right)}^{2018}}}}{9} + \dfrac{{{{\left( {3y + 0,4} \right)}^{2020}}}}{3}} \right] = 0\\
\Leftrightarrow {\left( {3x - 5} \right)^{2018}} + 3.{\left( {3y + 0,4} \right)^{2020}} = 0\\
{\left( {3x - 5} \right)^{2018}} \ge 0,\,\,\,\,\forall x\\
{\left( {3y + 0,4} \right)^{2020}} \ge 0,\,\,\,\forall y\\
\Rightarrow VT = {\left( {3x - 5} \right)^{2018}} + 3.{\left( {3y + 0,4} \right)^{2020}} \ge 0 = VP\,\,\,\,\,\,\forall x,y\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {3x - 5} \right)^{2018}} = 0\\
{\left( {3y + 0,4} \right)^{2020}} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x - 5 = 0\\
3y + 0,4 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3x = 5\\
3y = - 0,4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{5}{3}\\
y = - \dfrac{2}{{15}}
\end{array} \right.
\end{array}\)
Vậy \(x = \dfrac{5}{3};\,\,\,y = - \dfrac{2}{{15}}\)
