x/a+y/b=abx+1,x/b+y/a=aby+1,ab≠0,1/a+1/b≠ab,1/a-1/b≠ab,求xy的解
1个回答

x + ay = ab(abx+1),

ax + by = ab(aby + 1).

两式相减有（b-a)x - (b-a)y = ab(abx - aby)

即(b-a)(x-y) = (x-y)(ab)^2,所以（b-a - (ab)^2)(x-y) = 0.

由1/a-1/b≠ab,两边同乘以ab得：b-a≠(ab)^2,从而b-a - (ab)^2≠0.所以x-y = 0.

代入bx + ay = ab(abx+1),

中,有（a+b - （ab)^2)x = ab.由1/a+1/b≠ab,两边同乘以ab得：b+a≠(ab)^2,从而b+a - (ab)^2≠0,所以x=y=ab/[（a+b - （ab)^2)]

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