btsr1126
幼苗
共回答了23个问题采纳率:91.3% 举报
1/x(x+3)=1/3 * [1/x - 1/(x+3)],
1/(x+3)(x+6)=1/3 * [1/(x+3) - 1/(x+6)],
1/(x+6)(x+9)=1/3 * [1/(x+6) - 1/(x+9)],
所以左式
1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)
= 1/3 * [1/x - 1/(x+9)],
= 1/3 * (x+9-x) / x(x+9)
= 1/3 * 9/ x(x+9)
= 3/ x(x+9)
因为原式左右两边相等,则:
3/x(x+9) = 3/(2x+18)
==>x(x+9) = 2x + 18
==>x^2 + 7x - 18 = 0
==>x1 = 2,x2 = -9 .
因为x= -9代入右式得分母为0,所以舍去.
以 x= 2 代入原式检查:左边= 1/10 + 1/40 + 1/88 = 3/22
右边= 3/(4+18)= 3/22 ,左右两式相等.
1年前
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