若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.
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若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.(1+i)x+
若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.
若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.
若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.
(1+i)x+(1-i)y=2
x+y+i*(x-y)=2
从而有
x+y=2
x-y=0
联立解得
x=y=1
所以x*y=1
(1+i)x+(1-i)y=2
=>
X+Y+i*(X-Y)=2
=>
X+Y=2
X-Y=0
联立解得
X=Y=1 望采纳 谢谢~
∵(1+i)x+(1-i)y=2
x+xi+y-yi=2
(x+y)+(x-y)i=2+0i
∴x+y=2,x-y=0
∴x=y=1
故xy=1
若实数x,y满足(1+i)x+(1-i)y=2,求x*y的值.
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