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一道初中题——解方程组

解方程组
$\left\{ {\begin{array}{*{20}{c}}
  {\left( {y + c} \right)\left( {z + b} \right) = {a^2}} \\
  {\left( {z + a} \right)\left( {x + c} \right) = {b^2}} \\
  {\left( {x + b} \right)\left( {y + a} \right) = {c^2}}
\end{array}} \right.$
答案是\[\left\{\{x\to -b-c,y\to -a-c,z\to -a-b\},\left\{x\to \frac{-a b^2-a c^2+b^2 c+b c^2}{a b+a c-b c},y\to \frac{-a^2 b+a^2 c+a c^2-b c^2}{a b-a c+b c},z\to \frac{-a^2 b+a^2 c-a b^2+b^2 c}{a b-a c-b c}\right\}\right\}\]
我想了半天(1)+(2)-(3)等等也没出来。。

本帖最后由 hbghlyj 于 2019-10-17 22:09 编辑

回复 1# hbghlyj
法①从(1)(2)中解出x,y代入(3),$\left( {\frac{{{b^2}}}{{z + a}} - c + b} \right)\left( {\frac{{{a^2}}}{{z + b}} - c + a} \right) = {c^2}$
$\left( {\left( {{\text{b}} - {\text{c}}} \right){\text{z}} + {\text{ab}} - {\text{ac}} + {{\text{b}}^2}} \right)\left( {\left( {a - c} \right)z - bc + ab + {a^2}} \right) = {c^2}\left( {{\text{z}} + {\text{a}}} \right)\left( {{\text{z}} + {\text{b}}} \right)$
解得z=-a-b,$\frac{-a^2 b+a^2 c-a b^2+b^2 c}{a b-a c-b c}$
这种解法感觉太麻烦。破坏了轮换对称性。
法②观察得一组解(-b-c,-c-a,-a-b),除此之外,令x’=$\frac1{x+b+c}$,y’=$\frac1{y+c+a}$,z’=$\frac1{z+a+b}$,化为$\left\{ {\begin{array}{*{20}{c}}
  {\left( {\frac{1}{{y'}} - a} \right)\left( {\frac{1}{{z'}} - a} \right) = {a^2}} \\
  {\left( {\frac{1}{{z'}} - b} \right)\left( {\frac{1}{{x'}} - b} \right) = {b^2}} \\
  {\left( {\frac{1}{{x'}} - c} \right)\left( {\frac{1}{{y'}} - c} \right) = {c^2}}
\end{array}} \right.\left\{ {\begin{array}{*{20}{c}}
  {y' + z' = \frac{1}{a}} \\
  {z' + x' = \frac{1}{b}} \\
  {x' + y' = \frac{1}{c}}
\end{array}} \right.$
(x’,y’,z’)=$\left( {\frac{1}{2}\left( {\frac{1}{{\text{b}}} + \frac{1}{{\text{c}}} - \frac{1}{{\text{a}}}} \right),\frac{1}{2}\left( {\frac{1}{{\text{c}}} + \frac{1}{{\text{a}}} - \frac{1}{{\text{b}}}} \right)\frac{1}{2}\left( {\frac{1}{{\text{a}}} + \frac{1}{{\text{b}}} - \frac{1}{{\text{c}}}} \right)} \right)$
(x,y,z)=$\left( {\frac{{ - a{b^2} + {b^2}c - a{c^2} + b{c^2}}}{{ab + ac - bc}},\frac{{ - {a^2}b + {a^2}c + a{c^2} - b{c^2}}}{{ab - ac + bc}},\frac{{ - {a^2}b - a{b^2} + {a^2}c + {b^2}c}}{{ab - ac - bc}}} \right)$

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