题目内容
【题目】已知x、y、z为有理数,且|x+y+z+1|=x+y﹣z﹣2,则
=____________.
【答案】0
【解析】
根据绝对值的意义得到|x+y+z+1|=x+y+z+1或|x+y+z+1|=-(x+y+z+1),则x+y+z+1=x+y-z-2或-(x+y+z+1)=x+y-z-2,解得z=-
或x+y=
,然后把z=-
或x+y=
分别代入(x+y
)(2z+3)中计算即可.
∵|x+y+z+1|=x+y+z+1或|x+y+z+1|=-(x+y+z+1),
∴x+y+z+1=x+y-z-2或-(x+y+z+1)=x+y-z-2,
∴z=-
或x+y=
,
当z=-
时,(x+y
)(2z+3)=(x+y-
)[2×(-
)+3]=0;
当x+y=
时,(x+y
)(2z+3)=(
-
)(2z+3)=0,
综上所述,(x+y
)(2z+3)的值为0.
故答案为:0.
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