方程x^2+1=0和x^4+1=0在有理数集上是否同解?在实数集上呢?在复数集上呢? x^2+x+1=0的解？
x^4-10x^2 1=0\u5728\u6709\u7406\u6570\u57df\u4e0a\u53ef\u7ea6\u561b?x^4-10x^2 1=0
x^4\uff081-10x^17\uff09=0
x=0\u662f\u6709\u7406\u6839\uff0c\u6240\u4ee5
x^4-10x^2 1=0\u5728\u6709\u7406\u6570\u57df\u4e0a\u53ef\u7ea6
\u56e0\u4e3a \u5224\u522b\u5f0f b²-4ac=-3<0
\u6240\u4ee5 x^2+x+1=0\u5728\u5b9e\u6570\u8303\u56f4\u5185\u65e0\u89e3\uff0c
x^2+x+1=0\u5728\u590d\u6570\u8303\u56f4\u5185\u7684\u89e3\u662f\uff1ax=[-1\u00b1(\u6839\u53f73)i]/2\u3002
x^2+x+1=0...1\u5f0f
(\u56e0\u4e3ax\u22600)1\u5f0f\u9664\u4ee5x\u5f97:x+1+1/x=0...2\u5f0f
1\u5f0f-2\u5f0f\u5f97:x^2-1/x=0...3\u5f0f
3\u5f0f\u00d7x\u5f97:x^3-1=0 \u6240\u4ee5x=1
\u5f62\u5f0f\uff1a
\u628a\u76f8\u7b49\u7684\u5f0f\u5b50\uff08\u6216\u5b57\u6bcd\u8868\u793a\u7684\u6570\uff09\u901a\u8fc7\u201c=\u201d\u8fde\u63a5\u8d77\u6765\u3002
\u7b49\u5f0f\u5206\u4e3a\u542b\u6709\u672a\u77e5\u6570\u7684\u7b49\u5f0f\u548c\u4e0d\u542b\u672a\u77e5\u6570\u7684\u7b49\u5f0f\u3002
\u4f8b\u5982\uff1a
x+1=3\u2014\u2014\u542b\u6709\u672a\u77e5\u6570\u7684\u7b49\u5f0f\uff1b
2+1=3\u2014\u2014\u4e0d\u542b\u672a\u77e5\u6570\u7684\u7b49\u5f0f\u3002
\u9700\u8981\u6ce8\u610f\u7684\u662f\uff0c\u4e2a\u522b\u542b\u6709\u672a\u77e5\u6570\u7684\u7b49\u5f0f\u65e0\u89e3\uff0c\u4f46\u4ecd\u662f\u7b49\u5f0f\uff0c\u4f8b\u5982\uff1ax+1=x\u2014\u2014x\u65e0\u89e3\u3002
在复数集上：x²+1=0
x²=-1有两个根为x=i或x=-i
x⁴+1=0
x⁴=-1
x⁴=cosπ/2+isinπ/2
它有四个根：cos[(2kπ+π/2)/4]+isin[（）2kπ+π/2）/4] （k=0,1,2,3）0
解：两个方程在有理数集上同解，都是无解；在实数集上同解，都是无解；在复数集上不同解，方程x＾2+1＝0有两解，是i和-i，而方程x＾4＋1＝0有四解，是√2/2±i√2/2，-√2/2±i√2/2。
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