1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/(x+4)

鱼神tt2022-10-04 11:39:541条回答

1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/(x+4)
解方程

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ecosmos 共回答了21个问题 | 采纳率76.2%: 1/(x²+3x+2)=[(x+2)-(x+1)]/(x+1)(x+2)=1/(x+1) - 1/(x+2)
同理
1/(x²+5x+6)=1/(x+2) - 1/(x+3)
1/(x²+7x+12)=1/(x+3) - 1/(x+4)
所以方程成为 1/(x+1) - 1/(x+4) = 1/(x+4)
解得 x=2; 1年前

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原式=[1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)] ×(x+4)(x+5)
=[1/(x+1)-1/(x+2) +1/(x+2)-1/(x+3) +1/(x+3)-1/(x+4)]×(x+4)(x+5)
= [1/(x+1)-1/(x+4)]×(x+4)(x+5)
=3/(x+1)(x+4)×(x+4)(x+5)
=3(x+5)/(x+1)
当x=√2时
=3(√2+5)/(√2+1)
=3(√2+5)(√2-1)
=3(2-√2+5√2-5)
=12√2 -9

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