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數論的題目,高斯符號的題目,解聯立方程式

數論的題目,高斯符號的題目,解聯立方程式

設 \(a\) 為任意數,\([a]\) 表高斯符號,\((a )\) 表 \(a-[a]\) ,求下列聯立方程的解:
\[\left\{ {\begin{array}{*{20}{c}}
   {x + [y] + (z) = 1.5}  \;.............. (1)\\
   {y + [z] + (x) = 7.7}  \;.............. (2)\\
   {z + [x] + (y) = 2.6}  \;.............. (3)\\
\end{array}} \right.\]



解答:

由三式相加,可得
\[ x + y+ z + [x] + [y] + [z] + ( x) + ( y) + ( z) = 11.8\]
\[\Rightarrow 2(x+y+z) = 11.8 \Rightarrow  x+y+z=5.9\; ....... (4)\]

由 \((4)-(1),\, (4)-(2),\,(4)-(3)\),得
\[\left\{\begin{array}{*{20}{c}}
   y- [y] + z - (z ) = 4.4\\
   z- [z] + x- (x ) = -1.8\\
   x- [x] + y- (y ) = 3.3\\
\end{array}\right.\]
\[\Rightarrow \left\{\begin{array}{*{20}{c}}
   (y) + [z] = 0.4+4\\
   (z) + [x] = 0.2+\left(-2\right)\\
   (x) + [y] = 0.3+3\\
\end{array}\right.\]
\[\Rightarrow \left\{\begin{array}{*{20}{cc}}
   (y)=0.4,& [z] = 4\\
   (z)=0.2,& [x] = \left(-2\right)\\
   (x)=0.3,& [y] = 3\\
\end{array}\right.\]
\[\Rightarrow \left\{\begin{array}{*{20}{c}}
   x= (x) + [x] = -1.7 \\
   y=(y) + [y] = 3.4\\
   z=(z) + [z] = 4.2\\
\end{array}\right. .\]

多喝水。

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