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由題設:3a²;+5|b|=7可得: 3a²;=7-5|b|.且|b|≤7/5 ∴-19|b|≥-133/5 ∴14-19|b|≥-63/5 ∴3S =2(3a²;)-9|b| =14-10|b|-9|b| =14-19|b|≥-63/5 即恒有:3S≥-63/5 ∴S≥-21/5
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