关于x的方程1/x²+2x+10+1/x²+11x+10+1/x²-13x+10=0的解为

学习 时间:2026-06-05 18:50:09 阅读:1869
关于x的方程1/x²+2x+10+1/x²+11x+10+1/x²-13x+10=0的解为

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迷人的绿草

高大的项链

2026-06-05 18:50:09

写法容易引起歧义,是解1/(x²+2x+10)+1/(x²+11x+10)+1/(x²-13x+10) = 0吧?
∵1/(x²+2x+10)+1/(x²+11x+10)+1/(x²-13x+10) = 0,
∴1/(x+2+10/x)+1/(x+11+10/x)+1/(x-13+10/x) = x/(x²+2x+10)+x/(x²+11x+10)+x/(x²-13x+10) = 0。
换元y = x+10/x,可知1/(y+2)+1/(y+11)+1/(y-13) = 0。
∴3y²-147 = (y+11)(y-13)+(y+2)(y-13)+(y+2)(y+11) = 0,即有y² = 49。
解得y = ±7,代回x+10/x = y解得x = ±2,±5。
易验证它们都不是增根,因此原方程的解为x = ±2,±5。

最新回答共有2条回答

  • 怡然的过客
    回复
    2026-06-05 18:50:09

    写法容易引起歧义,是解1/(x²+2x+10)+1/(x²+11x+10)+1/(x²-13x+10) = 0吧?∵1/(x²+2x+10)+1/(x²+11x+10)+1/(x²-13x+10) = 0,∴1/(x+2+10/x)+1/(x+11+10/x)+1/(x-13+10/x) = x/(x²+2x+10)+x/(x²+11x+10)+x/(x²-13x+10) = 0。换元y = x+10/x,可知1/(y+2)+1/(y+11)+1/(y-13) = 0。∴3y²-147 = (y+11)(y-13)+(y+2)(y-13)+(y+2)(y+11) = 0,即有y² = 49。解得y = ±7,代回x+10/x = y解得x = ±2,±5。易验证它们都不是增根,因此原方程的解为x = ±2,±5。

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