已知(x+y一1)^2与丨x+2丨互为相反数,a,b互为倒数,则x^y+ab=
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已知(x+y一1)^2与丨x+2丨互为相反数,a,b互为倒数,则x^y+ab=
已知(x+y一1)^2与丨x+2丨互为相反数,a,b互为倒数,则x^y+ab=
已知(x+y一1)^2与丨x+2丨互为相反数,a,b互为倒数,则x^y+ab=
已知(x+y一1)^2与丨x+2丨互为相反数,
即
(x+y一1)^2+丨x+2丨=0
x+y-1=0,x+2=0
x=-2,y=1-x=1+2=3
又
a,b互为倒数,
ab=1
所以
x^y+ab
=(-2)^3+1
=-8+1
=-7