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若|x-1|+(y+2)2=0,则x-y=______

∵|x-1|+|y+2|+|z-3|=0,∴x-1=0,y+2=0,z-3=0,∴x=1,y=-2,z=3,∴(x+1)(y-2)(z-3)=(1+1)×(-2-2)×(3-3)=2×(-4)×0=0.故答案为0.

∵|x-1|+(y+2)2=0,∴x-1=0,y+2=0,∴x=1,y=-2.∴x-y=1-(-2)=1+2=3.

根据题意得,x-y+1=0,2+x=0,解得x=-2,y=-1,所以,x+y-xy=-2-1-(-2)×(-1)=-3-2=-5.故答案为:-5.

∵|x-1|+(y+2) 2 =0,∴x-1=0,y+2=0,解得x=1,y=-2,∴x+y=-1.故答案为:-1.

∵|x-1|+|y+2|=0,∴x=1,y=-2;原式=-(x+y)=-(1-2)=-1.故答案为-1.

∵|2x+y+1|+(x-y+2)2=0,∴2x+y+1=0x?y+2=0,解得:x=?1y=1,∴(xy?y)2y+x=-1.故答案为:-1.

(x-1)(y+2)=0的否定为(x-1)(y+2)≠0, x=1或y=-2的否定为x≠1且y≠-2.即命题“若(x-1)(y+2)=0,则x=1或y=-2”的否命题是:“若(x-1)(y+2)≠0,则x≠1,且y≠-2”.故答案为:若(x-1)(y+2)≠0,则x≠1,且y≠-2.

∵x、y满足|x-2|+(y+3) 2 =0,∴x-2=0,x=2;y+3=0,y=-3;则y x =(-3) 2 =9.

由题意得,x+2=0,y-2=0,解得x=-2,y=2,所以,(xy)2014=(?22)2014=1.故答案为:1.

∵(x+1)2+|y-1|=0,∴x+1=0,y-1=0,∴x=-1,y=1,∴x2008+y2009=1+1=2,故答案为2.

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