Đáp án:
\(\begin{array}{l}
a,\\
\left\{ \begin{array}{l}
x = 5\\
y = 3
\end{array} \right.\\
b,\\
{x^3} + {y^3} = 56
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^3} + {y^3} = 152\\
\Leftrightarrow \left( {x + y} \right).\left( {{x^2} - x.y + {y^2}} \right) = 152\\
\Leftrightarrow \left( {x + y} \right).19 = 152\\
\Leftrightarrow x + y = 8\\
\left\{ \begin{array}{l}
x + y = 8\\
x - y = 2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{\left( {x + y} \right) + \left( {x - y} \right)}}{2} = \dfrac{{8 + 2}}{2} = 5\\
y = \dfrac{{\left( {x + y} \right) - \left( {x - y} \right)}}{2} = \dfrac{{8 - 2}}{2} = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 5\\
y = 3
\end{array} \right.\\
b,\\
x + y = 2\\
\Leftrightarrow {\left( {x + y} \right)^2} = {2^2}\\
\Leftrightarrow {x^2} + 2xy + {y^2} = 4\\
\Leftrightarrow \left( {{x^2} + {y^2}} \right) + 2xy = 4\\
\Leftrightarrow 20 + 2xy = 4\\
\Leftrightarrow xy = - 8\\
{x^3} + {y^3} = \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\\
= \left( {x + y} \right).\left[ {\left( {{x^2} + {y^2}} \right) - xy} \right]\\
= 2.\left[ {20 - \left( { - 8} \right)} \right]\\
= 56\\
\Rightarrow {x^3} + {y^3} = 56
\end{array}\)
