若2x-5y+4z=17,2x+10y+z=13,则x+y+z=?帮个忙,OK?
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若2x-5y+4z=17,2x+10y+z=13,则x+y+z=?帮个忙,OK?若2x-5y+4z=17,2x+10y+z=13,则x+y+z=?帮个忙,OK?若2x-5y+4z=17,2x+10y+
若2x-5y+4z=17,2x+10y+z=13,则x+y+z=?帮个忙,OK?
若2x-5y+4z=17,2x+10y+z=13,则x+y+z=?帮个忙,OK?
若2x-5y+4z=17,2x+10y+z=13,则x+y+z=?帮个忙,OK?
在下认为此题有无数组解,以下证明:
两式相加,得4x+5(y+z)=30
则y+z=(30-4x)/5
∴x+y+z=x+(30-4x)/5=(30+x)/5
∵无论x取什么值,都有一组y、z满足题中方程
∴x可取任意值
即x+y+z可等于任意值
个人认为此题有误
第二个方程的x前系数应该为3
或者x、y、z有条件限制,比如为正整数等等.
希望LZ仔细看一下题目
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