Đáp án:
a. x=y
Giải thích các bước giải:
\(\begin{array}{l}
3)a.Có:{x^2} + {y^2} = 2xy\\
\to {x^2} + {y^2} - 2xy = 0\\
\to {\left( {x - y} \right)^2} = 0\\
\to x - y = 0\\
\to x = y\\
\to dpcm\\
b.\;{x^3} + {y^3} + {z^3} - 3xyz = 0\\
\to {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) + {z^3} - 3xyz\; = 0\\
\to {\left( {x + y} \right)^3} + {z^3} - 3xy\left( {x + y + z} \right)\; = 0\\
\to {\left( {x + y + z} \right)^3} - 3\left( {x + y + z} \right)\left( {x + y} \right)z - 3xy\left( {x + y + z} \right)\; = 0\\
\to {\left( {x + y + z} \right)^3} - 3\left( {x + y + z} \right)\left( {xy + yz + zx} \right)\; = 0\\
\to \left( {x + y + z} \right)\left[ {{{\left( {x + y + z} \right)}^2} - 3xy - 3yz - 3zx} \right]\;\\
\to \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right) = 0\left( 1 \right)\\
Do:x + y + z = 0\\
\left( 1 \right) \to 0.\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right) = 0\left( {ld} \right)\\
\to dpcm
\end{array}\)
