Question

解方程组:(x+y)^2-4(x+y)=5,(x-y)^2-2(x-y)=3.

Answer 1

(x+y)^2-4(x+y)=5,
(x-y)^2-2(x-y)=3.

(x+y)^2-4(x+y)-5=0
(x-y)^2-2(x-y)-3=0

(x+y-5)(x+y+1)=0
(x-y-3)(x-y+1)=0

x+y=5----------1)或x+y=-1----------2)
x-y=3----------3)或x-y=-1----------4)

联立1),3)解得:x=4,y=1
联立1),4)解得:x=2,y=3
联立2),3)解得:x=1,y=-2
联立2),4)解得:x=-1,y=0
以上四组均是方程组的解

Answer 2

由1式得
[(x+y)+1][(x+y)-5]=0
得x+y=-1或x+y=5
由2式得
[(x-y)+1][(x-y)-3]=0
得x-y=-1或x-y=3

①x+y=-1 x-y=-1
x=-1 y=0
②x+y=-1 x-y=3
x=1 y=-2
③x+y=5 x-y=-1
x=2 y=3
④x+y=5 x-y=3
x=4 y=1

综上，①x=-1 y=0 ②x=1 y=-2 ③x=2 y=3 ④=4 y=1

Answer 3

由1式得
[(x+y)+1][(x+y)-5]=0
得x+y=-1,x+y=5
由2式得
[(x-y)+1][(x-y)-3]=0
得x-y=-1,x-y=3
综合得
x+y=5
x-y=3
得x=4,y=1