Toán Lớp 9: √x-2 + √y-3 + √z-5 = $\frac{1}{2}$ (x+y+z-7)
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Toán Lớp 9: √x-2 + √y-3 + √z-5 = $\frac{1}{2}$ (x+y+z-7)
Comments ( 1 )
Dkxd:\left\{ \begin{array}{l}
x \ge 2\\
y \ge 2\\
z \ge 5
\end{array} \right.\\
\sqrt {x – 2} + \sqrt {y – 3} + \sqrt {z – 5} = \frac{1}{2}\left( {x + y + z – 7} \right)\\
\Leftrightarrow x + y + z – 7 = 2\sqrt {x – 2} + 2\sqrt {y – 3} + 2\sqrt {z – 5} \\
\Leftrightarrow x – 2 – 2\sqrt {x – 2} + 1 + y – 3 – 2\sqrt {y – 3} + 1\\
+ z – 5 – 2\sqrt {z – 5} + 1 = 0\\
\Leftrightarrow {\left( {\sqrt {x – 2} – 1} \right)^2} + {\left( {\sqrt {y – 3} – 1} \right)^2} + {\left( {\sqrt {z – 5} – 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {x – 2} – 1 = 0\\
\sqrt {y – 3} – 1 = 0\\
\sqrt {z – 5} – 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {x – 2} = 1\\
\sqrt {y – 3} = 1\\
\sqrt {z – 5} = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x – 2 = 1\\
y – 3 = 1\\
z – 5 = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 3\left( {tm} \right)\\
y = 4\left( {tm} \right)\\
z = 6\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 3;y = 4;z = 6
\end{array}$