x²+4x+3=(x+1)(x+3)
x³+27 =(x+3)(x²-3x+9)
x²-3x+2=(x-1)(x-3)
∵(x-1) / (x²+4x+3) 与 (x³+27) / (x²-3x+2) 互为倒数
∴[(x-1) / (x²+4x+3)] × [(x³+27) / (x²-3x+2)] = 1
∴(x-1)(x³+27) = (x²+4x+3)(x²-3x+2)
∴(x-1)(x+3)(x²-3x+9) = (x+1)(x+3)(x-1)(x-3)
∴x²-3x+9 = (x+1)(x-3)
∴x²-3x+9 = x²-2x-3
∴x=12
x为何值时代数式x-1/x2+4x+3与x3+27/x2-3x+2互为倒数
x为何值时代数式x-1/x2+4x+3与x3+27/x2-3x+2互为倒数
数学人气:975 ℃时间:2019-11-02 13:39:59
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