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若x、y、z满足3x+7y+z=1和4x+10y+z=2001,则分式
2000x+2000y+2000z
x+3y
的值为______.
人气:277 ℃ 时间:2020-04-14 03:26:32
解答
由x、y、z满足3x+7y+z=1和4x+10y+z=2001,
得出:
2(x+3y)+(x+y+z)=1
3(x+3y)+(x+y+z)=2001
,解得:
x+3y=2000
x+y+z=−3999

2000x+2000y+2000z
x+3y
=
2000(x+y+z)
x+3y

=
2000×(−3999)
2000
=-3999.
故答案为:-3999.
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