已知x+y=-2,求x∧2-4x+y∧2+2xy-4y+4的值RT
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已知x+y=-2,求x∧2-4x+y∧2+2xy-4y+4的值RT
已知x+y=-2,求x∧2-4x+y∧2+2xy-4y+4的值
RT
已知x+y=-2,求x∧2-4x+y∧2+2xy-4y+4的值RT
原式=(x+y)^2-4(x+y)+3=4+8+3=15
原式=(X+Y)^2-4(X+Y)+4因为X+Y=-2所以带入得(-2)^2-4*(-2)+4最后答案为0采纳吧!
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