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∵在△ABC中,3a2+3b2-3c2+2ab=0,即a2+b2-c2=-23ab,∴由余弦定理得cosC=a2+b2−c22ab=−23ab2ab=-13<0,即C為鈍角,∴sinC=1−cos2C=223則tanC=sinCcosC=-22.故答案為:-22
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