摘要：已知:|x-1|≤1, 求证:(1)|2x+3|≤7; (2)|x2-1|≤3 证明:(1)∵|2x+3|=|2(x-1)+5|≤2|x-1|+5≤2+5=7 (2)|x2-1|=|(x+1)(x-1)|=|(x-1)[(x-1)+2]| ≤|x-1||(x-1)+2|≤|x-1|+2≤1+2=3

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