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三道初中题——解方程组

本帖最后由 hbghlyj 于 2020-2-12 10:34 编辑

(1)给定非零实数A,解方程组\[\left\{ \begin{gathered}
  (x - y)(y + z) = 1 - {A^2} \hfill \\
  (y - z)(x + z) =  - 1 - \frac{1}{{{A^2}}} \hfill \\
  (z - x)(x + y) =  - 1 \hfill \\
\end{gathered}  \right.\]
机器解出$(x,y,z)=\left( {0,A, - \frac{1}{A}} \right)\left( {0, - A,\frac{1}{A}} \right)\left( {\frac{{{A^3} + A - 1}}{{2A\sqrt {{A^2} + A + 1} }}, - \frac{{{A^3} - A - 1}}{{2A\sqrt {{A^2} + A + 1} }}, - \frac{{{A^3} + 2{A^2} + A + 1}}{{2A\sqrt {{A^2} + A + 1} }}} \right)\left( { - \frac{{{A^3} + A - 1}}{{2A\sqrt {{A^2} + A + 1} }},\frac{{{A^3} - A - 1}}{{2A\sqrt {{A^2} + A + 1} }},\frac{{{A^3} + 2{A^2} + A + 1}}{{2A\sqrt {{A^2} + A + 1} }}} \right)\left( {\frac{{{A^3} + A + 1}}{{2A\sqrt {{A^2} - A + 1} }}, - \frac{{{A^3} - A + 1}}{{2A\sqrt {{A^2} - A + 1} }}, - \frac{{{A^3} - 2{A^2} + A - 1}}{{2A\sqrt {{A^2} - A + 1} }}} \right)\left( { - \frac{{{A^3} + A + 1}}{{2A\sqrt {{A^2} - A + 1} }},\frac{{{A^3} - A + 1}}{{2A\sqrt {{A^2} - A + 1} }},\frac{{{A^3} - 2{A^2} + A - 1}}{{2A\sqrt {{A^2} - A + 1} }}} \right)$
(2)给定$\ge2$的整数n,正数A,解方程组$x_1^i+x_2^i+\cdots+x_n^i=A,i=1,2,\cdots,n$
(3)给定$\ge2$的整数n,正数A,解方程组$x_1^i+x_2^i+\cdots+x_n^i=A^i,i=1,2,\cdots,n$

本帖最后由 hbghlyj 于 2020-2-12 10:28 编辑

下面是我的解法,可能不太漂亮..
(1)令x'=x-y,y'=y-z,z'=y+z,化为
\[\left\{ \begin{gathered}
  x'\left( {x' + z'} \right) = 1①\\
  x{'^2} + y'z' = {A^2}②\\
  y'\left( { - x' + y' - z'} \right) = 1 + \frac{1}{{{A^2}}}③\\
\end{gathered}  \right.\]
由①②解出\[\left\{ \begin{array}{l}
z' = \frac{1}{{x'}} - x'\\
y' = \frac{{x'\left( {{A^2} - x{'^2}} \right)}}{{1 - x{'^2}}}
\end{array} \right.\]代入③,$(A x'-1) (A x'+1) \left(x'^2-A^2+A-1\right) \left(x'^2-A^2-A-1\right)=0$

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