ï¼k-1ï¼ï¼X2+1ï¼+2X=0ï¼
åºè¯¥æ¯(k-1)(x²+1)+2x=0å§ï¼
解ï¼
(k-1)(x²+1)+2x=0ï¼è¯´æï¼æé¢
(k-1)x²+k-1+2x=0ï¼è¯´æï¼ä¹æ³åé
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(k-1)x²+2x+k-1=0ï¼è¯´æï¼ç§»é¡¹
1ãå½k=1æ¶ï¼æï¼
2x=0ï¼è¯´æï¼å°k=1代å
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解å¾ï¼x=0ï¼è¯´æï¼çå¼ä¸¤è¾¹åé¤ä»¥2ï¼çå¼ä¸åã
2ãå½kâ 1æ¶ï¼æï¼
x²+[2/(k-1)]x+1=0ï¼è¯´æï¼çå¼ä¸¤è¾¹åé¤ä»¥(k-1)
x²+2Ã[1/(k-1)]x+1=0ï¼è¯´æï¼å°2/(k-1)ï¼å解为2Ã[1/(k-1)]
x²+2Ã[1/(k-1)]x+[1/(k-1)]²-[1/(k-1)]²+1=0ï¼è¯´æï¼åæ¶å åä¸ä¸ªç¸ççå¼åï¼åå¼ä¸å
{x²+2Ã[1/(k-1)]x+[1/(k-1)]²}-[1/(k-1)]²+1=0ï¼è¯´æï¼å¿ä¸å°å¤é¡¹å¼åç»
[x+1/(k-1)]²-[1/(k-1)]²+1=0ï¼è¯´æï¼å©ç¨å®å
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[x+1/(k-1)]²=[1/(k-1)]²-1ï¼è¯´æï¼ç§»é¡¹
[x+1/(k-1)]²=1/(k-1)²-(k-1)²/(k-1)²ï¼è¯´æï¼çå¼å³è¾¹éå
[x+1/(k-1)]²={1-(k-1)²}/(k-1)²ï¼è¯´æï¼çå¼å³è¾¹è®¡ç®
[x+1/(k-1)]²=k(2-k)/(k-1)²ï¼è¯´æï¼å¯¹å³è¾¹ååè¿è¡è®¡ç®
x+1/(k-1)=±â[k(2-k)/(k-1)²]ï¼è¯´æï¼çå¼ä¸¤è¾¹åæ¶å¼å¹³æ¹
(1)ãå½2â¥kï¼1æ¶ï¼
x+1/(k-1)=±{â[k(2-k)]}/(k-1)ï¼è¯´æï¼å³è¾¹è¿è¡å¼å¹³æ¹è®¡ç®
x={-1±â[k(2-k)]}/(k-1)ï¼è¯´æï¼ç§»é¡¹å¹¶è®¡ç®ã
å¾å°æ¹ç¨ç解ï¼x1={-1+â[k(2-k)]}/(k-1)ï¼x2=-{1+â[k(2-k)]}/(k-1)ï¼
(2)ãå½kï¼2æ¶ï¼
x+1/(k-1)=±{â[k(2-k)]}/(k-1)ï¼è¯´æï¼å³è¾¹è¿è¡å¼å¹³æ¹è®¡ç®
æ¤æ¶ï¼k(2-k)ï¼0ï¼æ¹ç¨æ å®æ°è§£ï¼ä½æå¤æ°è§£
å¤æ°è§£æ¯ï¼x={-1±iâ[k(k-2)]}/(k-1)ï¼è¯´æï¼ç§»é¡¹å¹¶è®¡ç®ã
å¾å°æ¹ç¨ç解æ¯ï¼x1={-1+iâ[k(k-2)]}/(k-1)ï¼x2=-{1+iâ[k(k-2)]}/(k-1)ï¼
(3)ãå½0ï¼kï¼1æ¶ï¼
x+1/(k-1)=±{â[k(2-k)]}/(1-k)ï¼è¯´æï¼å³è¾¹è¿è¡å¼å¹³æ¹è®¡ç®ï¼æ³¨æï¼k-1ï¼0
x={1±â[k(k-2)]}/(1-k)ï¼è¯´æï¼ç§»é¡¹å¹¶è®¡ç®ã
å¾å°æ¹ç¨ç解æ¯ï¼x1={1+â[k(k-2)]}/(1-k)ï¼x2={1-â[k(k-2)]}/(1-k)ï¼
(4)ãå½kï¼0æ¶ï¼
x+1/(k-1)=±{â[k(2-k)]}/(1-k)ï¼è¯´æï¼å³è¾¹è¿è¡å¼å¹³æ¹è®¡ç®ï¼æ³¨æï¼k-1ï¼0
æ¤æ¶ï¼k(2-k)ï¼0ï¼æ¹ç¨æ å®æ°è§£ï¼ä½æå¤æ°è§£
å¤æ°è§£æ¯ï¼x={1±iâ[-k(2-k)]}/(1-k)ï¼è¯´æï¼ç§»é¡¹å¹¶è®¡ç®ã
å¾å°æ¹ç¨ç解æ¯ï¼x1={1+iâ[k(2-k)]}/(1-k)ï¼x2={1-iâ[k(2-k)]}/(1-k)ï¼
综ä¸æè¿°ï¼æç»æ¹ç¨ç解æ¯ï¼
1ãå½kâ¤0æ¶ï¼x1={1+iâ[k(2-k)]}/(1-k)ï¼x2={1-iâ[k(2-k)]}/(1-k)ï¼
2ãå½0ï¼kï¼1æ¶ï¼x1={1+â[k(k-2)]}/(1-k)ï¼x2={1-â[k(k-2)]}/(1-k)ï¼
3ãå½k=1æ¶ï¼x=0ï¼
4ãå½1ï¼kâ¤2æ¶ï¼x1={-1+â[k(2-k)]}/(k-1)ï¼x2=-{1+â[k(2-k)]}/(k-1)ï¼
5ãå½kï¼2æ¶ï¼x1={-1+iâ[k(k-2)]}/(k-1)ï¼x2=-{1+iâ[k(k-2)]}/(k-1)ã
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