若a
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若a
若a
若a
1.由a
a<0,且ab<0
则b>0
|b-a+4|-|a-b-7|
=b-a+4-[-(a-b-7)]
=b-a+4+(a-b-7)
=b-a+4+a-b-7
=-3
(3x-2y)^2+|x+2|=0
|x+2|=0,(3x-2y)^2=0
x=-2,(3x-2y)^2=0
(-2*3-2y)^2=0
(-6-2y)^2=0
y=-3
(3x+3y)/(xy-1)
=3(x+y)/(xy-1)
=3*(-2-3)/[(-2)*(-3)-1]
=-15/(6-1)
=-15/5
=-3
|b-a+4|-|a-b-7| =b-a+4-(7+b-a)=b-a+4-7-b+a=-3
因为a<0,且ab<0,所以,b>0.
|b-a+4|-|a-b-7| =b-a+4-(7+b-a)=b-a+4-7-b+a=-3
{b-a+4}-{b-a+7}=-3
-3