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题目

解方程

(x2+2x1)2+2x2+3x=3(x^2+2x-1)^2+2x^2+3x=3

解答

别问,问就是拉马努金

鬼知道怎么凑出来的,就只能算算看了

其实面对这种题,可以考虑先代入 1,-1, 0 等值试一下,如果能找到一个根,就直接使用长除法降次就好了…

(x2+2x1)2+2x2+3x=3(x2+2x1)2+2x2+3x3=0(x2+2x1)2+2(x2+2x1)+1x2=0(x2+2x1+1)2x2=0(x2+2x)2(x+2)=0x2(x+2)2(x+2)=0(x+2)(x2(x+2)1)=0(x+2)(x3+2x21)=0(x+2)(x3+2x2+x(x+1))=0(x+2)(x(x+1)2(x+1))=0(x+2)((x+1)(x2+x1))=0(x+2)(x+1)(x2+x1)=0\begin{align} (x^2+2x-1)^2+2x^2+3x &= 3 \\ (x^2+2x-1)^2+2x^2+3x-3 &= 0 \\ (x^2+2x-1)^2+2(x^2+2x-1)+1-x-2 &=0 \\ (x^2+2x-1+1)^2 -x -2 &=0 \\ (x^2+2x)^2 -(x+2) &=0 \\ x^2(x+2)^2 -(x+2) &=0 \\ (x+2)(x^2(x+2)-1) &=0 \\ (x+2)(x^3+2x^2-1) &=0 \\ (x+2)(x^3+2x^2 +x -(x+1)) &=0 \\ (x+2)(x(x+1)^2 -(x+1)) &=0 \\ (x+2)((x+1)(x^2+x-1)) &=0 \\ (x+2)(x+1)(x^2+x-1) &=0 \\ \end{align}

第六行之前可以换一种方式

(x2+2x1)2+2x2+3x=3x4+2x3+4x22xx22x+1+2x2+3x3=0x4+4x3+4x2x2=0x2(x+2)2(x+2)=0(x+2)(x2(x+2)1)=0\begin{align*} (x^2+2x-1)^2+2x^2+3x &= 3 \\ x^4+2x^3+4x^2-2x-x^2-2x+1+2x^2+3x-3 &=0 \\ x^4+4x^3+4x^2-x-2 &=0 \\ x^2(x+2)^2-(x+2) &= 0 \\ (x+2)(x^2(x+2)-1) &= 0 \end{align*}