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已知a与b互为倒数,c与d互为相反数,m为最大的负整数,n的绝对值为2,m/2+(3c+3d-25)/(4ab-2)=?
sorry,是m/2+(3c+3d-25)/(4ab-2)+mn(打错了)
人气:240 ℃ 时间:2019-10-11 04:23:36
解答
已知a与b互为倒数,ab=1,
c与d互为相反数,c+d=0;
m为最大的负整数,m=-1;
n的绝对值为2,n1=2,n2=-2,
m/2+(3c+3d-25)/(4ab-2)
=-1/2+[3(c+d))-25]/(4-2)
=-1/2+(-25/2)
=-26/2
=-13
[没有n什么事?]sorry,是m/2+(3c+3d-25)/(4ab-2)+mnm/2+(3c+3d-25)/(4ab-2)+mn1=-1/2+[3(c+d))-25]/(4-2)+(-1/2)*2=-1/2+(-25/2)-1=-26/2-1=-14m/2+(3c+3d-25)/(4ab-2)+mn2=-1/2+[3(c+d))-25]/(4-2)+(-1/2)*(-2)=-1/2+(-25/2)+1=-26/2+1=-12
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