若(X+Y)(X-6+Y)=-9,求2X+2Y+2002的值.(有分数送)

设(X+Y)=t

t*(t-6)=-9
t^2-6t+9=0
(t-3)^2=0
t=3

2X+2Y+2002
=2t+2002
=2008

设X+Y=U

U(U-6)=-9
U^2-6U+9=0
(U-3)^2=0
U=3

X+Y=3

2X+2Y+2002=2(X+Y)+2002=2(3)+2002=2008

把x+y看成一个整体a
就变成a*(a-6)=-9
解得a=3
要求的是2a+2002
=2008

设a=X+Y
原式变a(a-6)=-9
解得a=3
2X+2Y+2002=2a+2002=2008

设 X+Y=Z
Z(Z-6)=-9
Z=3
2X+2Y+2002=2Z+2002=2008

x+y看成一整体可求得x+y=3,故原式=2008