1.(2/2x-1) = (4/4x^2-1)2.(2x+1)/(x^2+x) = 5/(6x+6)
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1.(2/2x-1)=(4/4x^2-1)2.(2x+1)/(x^2+x)=5/(6x+6)1.(2/2x-1)=(4/4x^2-1)2.(2x+1)/(x^2+x)=5/(6x+6)1.(2/2x-
1.(2/2x-1) = (4/4x^2-1)2.(2x+1)/(x^2+x) = 5/(6x+6)
1.(2/2x-1) = (4/4x^2-1)
2.(2x+1)/(x^2+x) = 5/(6x+6)
1.(2/2x-1) = (4/4x^2-1)2.(2x+1)/(x^2+x) = 5/(6x+6)
1题:根据比例关系分子之比为2:4,则分母之比为1:2,所以2x-1=2(4x^2-1),然后解一元二次方程8x^2-2x-1=0
即(2x-1)(4x+1)=0
解得x1=1/2,x2=1/4
2题,交叉相乘,即(2x+1)(6x+6)=5(x^2+x)
化简得:7x^2+13x+6=0
即(7x+6)(x+1)=0
得x1=-6/7,x2=-1
1。x=0、5
2。x1=-1 x2=-6\7
1.用公式计算:COSX=1-X*X/2!X*X*X*X/4!-X*X*X*X*X*X/6!.直到最后一项的绝对
*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X ||
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一个“整式的乘法”的问题请先阅读下列解题过程,再仿做下面的问题.已知X*X + X - 1=0,求X*X*X + 2*X*X + 3的值.解X*X*X + 2X*X +3=X*X*X +X*X -X +X*X +X +3=X{X*X +X -1} +X*X +X -1 +4=0+0+4=4+ x + X*X + X*X*X=0.+ X*X + X*X*X
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x+2/x+1-x+4/x+3=x+6/x+5-x+8/x+7 x=?
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(x-2)(x+1)(x+4)(x+7)=19
(x+1)(x+2)(x+3)(x+4)=24