Question

Đã trả lời bởi chuyên gia

Answer 1

$\begin{array}{l}
{x^2} + {y^2} + {z^2} = xy + yz + zx\\
\Leftrightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2.\left( {xy + yz + zx} \right)\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} = 2xy + 2yz + 2zx\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx = 0\\
\Leftrightarrow \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{y^2} - 2yz + {z^2}} \right)\\
+ \left( {{z^2} - 2zx + {x^2}} \right) = 0\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x - y} \right)^2} \ge 0\\
{\left( {y - z} \right)^2} \ge 0\\
{\left( {z - x} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} \ge 0\\
Khi:{\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {y - z} \right)^2} = 0\\
{\left( {z - x} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
y - z = 0\\
z - x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y\\
y = z\\
z = x
\end{array} \right.\\
\Leftrightarrow x = y = z
\end{array}$

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Answer 2

$\begin{array}{l}
{x^2} + {y^2} + {z^2} = xy + yz + zx\\
\Leftrightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2.\left( {xy + yz + zx} \right)\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} = 2xy + 2yz + 2zx\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx = 0\\
\Leftrightarrow \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{y^2} - 2yz + {z^2}} \right)\\
+ \left( {{z^2} - 2zx + {x^2}} \right) = 0\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x - y} \right)^2} \ge 0\\
{\left( {y - z} \right)^2} \ge 0\\
{\left( {z - x} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} \ge 0\\
Khi:{\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {y - z} \right)^2} = 0\\
{\left( {z - x} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
y - z = 0\\
z - x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y\\
y = z\\
z = x
\end{array} \right.\\
\Leftrightarrow x = y = z
\end{array}$

Answer 3

$\begin{array}{l}
{x^2} + {y^2} + {z^2} = xy + yz + zx\\
\Leftrightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2.\left( {xy + yz + zx} \right)\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} = 2xy + 2yz + 2zx\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx = 0\\
\Leftrightarrow \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{y^2} - 2yz + {z^2}} \right)\\
+ \left( {{z^2} - 2zx + {x^2}} \right) = 0\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x - y} \right)^2} \ge 0\\
{\left( {y - z} \right)^2} \ge 0\\
{\left( {z - x} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} \ge 0\\
Khi:{\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {y - z} \right)^2} = 0\\
{\left( {z - x} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
y - z = 0\\
z - x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y\\
y = z\\
z = x
\end{array} \right.\\
\Leftrightarrow x = y = z
\end{array}$

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Answer 4

$\begin{array}{l}
{x^2} + {y^2} + {z^2} = xy + yz + zx\\
\Leftrightarrow 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2.\left( {xy + yz + zx} \right)\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} = 2xy + 2yz + 2zx\\
\Leftrightarrow 2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2zx = 0\\
\Leftrightarrow \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{y^2} - 2yz + {z^2}} \right)\\
+ \left( {{z^2} - 2zx + {x^2}} \right) = 0\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x - y} \right)^2} \ge 0\\
{\left( {y - z} \right)^2} \ge 0\\
{\left( {z - x} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow {\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} \ge 0\\
Khi:{\left( {x - y} \right)^2} + {\left( {y - z} \right)^2} + {\left( {z - x} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {y - z} \right)^2} = 0\\
{\left( {z - x} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
y - z = 0\\
z - x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y\\
y = z\\
z = x
\end{array} \right.\\
\Leftrightarrow x = y = z
\end{array}$

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