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设甲,乙,丙分别为x,x-1,x-2
则乙的倒数为:1/(x-1)
甲的倒数为:1/x
丙的倒数:1/(x-2)
于是,根据条件“乙数的倒数与甲数的倒数的2倍的和,与丙数的倒数的3倍相等”,可列出方程:
[1/(x-1) + (1/x)*2] = 3*[1/(x-2)]
1/(x-1) + 2/x = 2/(x-2)
左右两侧同时乘以"(x-1)*x*(x-2)":
x*(x-2) + 2*(x-1)*(x-2) = 3* x*(x-1)
x^-2x +2(x^-3x+2)=3(x^-x)
x^-2x +2x^-6x+4=3x^-3x
右侧移到左侧,并合并同类项:
x^-2x+2x^-6x+4-3x^+3x=0
(x^+2x^-3x^) +(3x-2x-6x) +4=0
0 - 5x +4=0
5x=4
x=4/5
于是,x-1=4/5 -1=-1/5
x-2=4/5 -2=-6/5
三个数分别是:
甲:4/5
乙:-1/5
丙:-6/5
1年前
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