已知x+1/x=2+1/2的解是x1=2,x2=1/2 x+1/x=3+1/3的解是x1=3,x2=1/3……请根据上面的规律吧关于x的方程(x...
已知x+1/x=2+1/2的解是x1=2,x2=1/2
x+1/x=3+1/3的解是x1=3,x2=1/3……请根据上面的规律吧关于x的方程(x的平方-x+1)/(x-1)=a+1/(a-1)变形为方程x+1/x=c+1/c的形式（ ），方程的解是（ ）写出解题过程。
解
ã1ã
x²-x+1
=(x²-2x+1)+x
=(x-1)²+x
=(x-1)²+(x-1)+1
å³ï¼x²-x+1=(x-1)²+(x-1)+1
â´(x²-x+1)/(x-1)=[(x-1)²+(x-1)+1]/(x-1)=(x-1)+1+[1/(x-1)]
å³åæ¹ç¨å·¦è¾¹=ï¼x-1)+1+[1/(x-1)]
åæ¹ç¨å¯å为ï¼
(x-1)+1+[1/(x-1)]=a+[1/(a-1)]
移项å¯å¾ï¼
(x-1)+[1/(x-1)]=(a-1)+[1/(a-1)]
令x-1=y, a-1=c. ä¸é¢æ¹ç¨å¯å为ï¼
y+(1/y)=c+(1/c)
ã2ã
æ¤æ¶ï¼æ¹ç¨ï¼
(x-1)+[1/(x-1)]=(a-1)+[1/(a-1)]ç解为ï¼
x₁-1=a-1 x₂-1=1/(a-1)
å³ï¼x₁=a. x₂=a/(a-1)
ã1ã
x²-x+1
=(x²-2x+1)+x
=(x-1)²+x
=(x-1)²+(x-1)+1
å³ï¼x²-x+1=(x-1)²+(x-1)+1
â´(x²-x+1)/(x-1)=[(x-1)²+(x-1)+1]/(x-1)=(x-1)+1+[1/(x-1)]
å³åæ¹ç¨å·¦è¾¹=ï¼x-1)+1+[1/(x-1)]
åæ¹ç¨å¯å为ï¼
(x-1)+1+[1/(x-1)]=a+[1/(a-1)]
移项å¯å¾ï¼
(x-1)+[1/(x-1)]=(a-1)+[1/(a-1)]
令x-1=y, a-1=c. ä¸é¢æ¹ç¨å¯å为ï¼
y+(1/y)=c+(1/c)
ã2ã
æ¤æ¶ï¼æ¹ç¨ï¼
(x-1)+[1/(x-1)]=(a-1)+[1/(a-1)]ç解为ï¼
x₁-1=a-1 x₂-1=1/(a-1)
å³ï¼x₁=a. x₂=a/(a-1)
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第1个回答 2012-07-29
(x^2-1+1)/(x-1)=(x^2-x)/(x-1)+1/(x-1)=x+/(x-1)
于是(x^2-1+1)/(x-1)=a+1/(a-1)变形为
x+1/(x-1)=a+1/(a-1)
即
(x-1)+1/(x-1)=(a-1)+1/(a-1)
这就是第一问的答案。方程解是x-1=a-1或x-1=1/(a-1)
整理的方程解为x=a或x=1+1/(a-1)
于是(x^2-1+1)/(x-1)=a+1/(a-1)变形为
x+1/(x-1)=a+1/(a-1)
即
(x-1)+1/(x-1)=(a-1)+1/(a-1)
这就是第一问的答案。方程解是x-1=a-1或x-1=1/(a-1)
整理的方程解为x=a或x=1+1/(a-1)
第2个回答 2012-07-29
(x²-x+1)/(x-1)=x+1/(x-1)，方程(x²-x+1)/(x-1)=a+1/(a-1)两边都减1得
x-1+1/(x-1)=a-1+1/(a-1)，方程的解为a-1
x-1+1/(x-1)=a-1+1/(a-1)，方程的解为a-1
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