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已知方程組,3x=y+3…①2kx−(k−1)y=6…②,由3x=y+3,移項得,y=3x-3,把y代入方程②得,2kx-(k-1)(3x-3)=6,∴x=9−3k3−k,把x值代入①得,y=18−6k3−k,(k≠3)∵x+y=0,∴9−3k3−k+18−6k3−k=0解得k=3.
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