本帖最后由 hbghlyj 于 2019-9-20 22:49 编辑
f(${{\text{x}}_1},{x_2}, \cdots ,{x_n}$)为多项式函数,求证：$f(\sqrt {{{\text{x}}_1}} ,\sqrt {{x_2}} , \cdots ,\sqrt {{x_n}} )$的有理化为 $\mathop \sum \limits_{\matrix
{{{\text{i}}_1} = \left| {{i_2}} \right| = \cdots = \left| {{{\text{i}}_{\text{n}}}} \right| = 1} \\
} {\text{f}}\left( {{{\text{i}}_1}\sqrt {{{\text{x}}_1}} ,{{\text{i}}_2}\sqrt {{{\text{x}}_2}} , \cdots ,{{\text{i}}_{\text{n}}}\sqrt {{{\text{x}}_{\text{n}}}} } \right)$.
举例如下
求$1 + \sqrt {\text{x}} + \sqrt {\text{y}} +2 \sqrt {{\text{xy}}} $的有理化因式
按照程序化作法：
将$1 + \sqrt {\text{y}} + \sqrt {\text{x}} \left( {1 + 2\sqrt y } \right)$乘以$1 + \sqrt {\text{y}} - \sqrt {\text{x}} \left( {1 + 2\sqrt y } \right)$得$1 - x + y - 4xy + 2\sqrt {\text{y}} \left( {1 - 2x} \right)$
乘以$1 - x + y - 4xy - 2\sqrt {\text{y}} \left( {1 - 2x} \right)$得${\left( {1 - {\text{x}} + {\text{y}} - 4{\text{xy}}} \right)^2} - 4{\text{y}}{\left( {1 - 2{\text{x}}} \right)^2}$是有理式，所以有理化因式为$\left( {1 + \sqrt {\text{y}} - \sqrt {\text{x}} \left( {1 + 2\sqrt y } \right)} \right)\left( {1 - {\text{x}} + {\text{y}} - 4{\text{xy}} - 2\sqrt {\text{y}} \left( {1 - 2x} \right)} \right)$
但是按照定理可直接写出：有理化为$\left(2 \sqrt{x y}+\sqrt{x}+\sqrt{y}+1\right) \left(-2 \sqrt{x y}-\sqrt{x}+\sqrt{y}+1\right) \left(-2 \sqrt{x y}+\sqrt{x}-\sqrt{y}+1\right) \left(2 \sqrt{x y}-\sqrt{x}-\sqrt{y}+1\right)$
有理化因式为$\left(-2 \sqrt{x y}-\sqrt{x}+\sqrt{y}+1\right) \left(-2 \sqrt{x y}+\sqrt{x}-\sqrt{y}+1\right) \left(2 \sqrt{x y}-\sqrt{x}-\sqrt{y}+1\right)$ |
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发表于 2019-9-20 13:11
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发表于 2019-9-24 16:42
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