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log2(x^2-3x)=log二分之一(x^2-x+6)
人气:161 ℃ 时间:2020-02-05 16:36:17
解答
log2(x²-3x)-log1/2(x^2-x+6)=0log2(x²-3x)-[log2(x²-x+6)]/log2 1/2=0log2(x²-3x)+log2(x²-x+6)=0log2(x²-3x)(x²-x+6)=0(x²-3x)(x²-x+6)=1剩下自己解...
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