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①x(x+1)(x+2)(x+3)=3
﹙x²+3x﹚﹙x²+3x+2﹚=3 ﹙x²+3x﹚²+2﹙x²+3x﹚-3=0 ﹙x²+3x+3﹚﹙x²+3x-1﹚=0 则x²+3x+3=0（无实根，舍）或x²+3x-1=0 解得：x=[-3±√﹙13﹚]÷2 ②x^4+8x^3+14x^2+8x+1=0 ﹙x^4-2x^2+1﹚+8x(x^2+2x+1)=0 (x+1)²(x-1)²+8x(x+1)²=0 (x+1)²(x²+6x+1)=0 ∴(x+1)²=0或x²+6x+1=0 ∴x1=x2=-1,x3=-3+2√2,x4=-3-2√2 ③x(x+1)(x^2-2x-4)=0 ∴x=0,x+1=0,x^2-2x-4=0 ∴x1=0,x2=-1,x3=1+√5,x4=1-√5 |
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