Maths

√(x-1) + (xy-3)^2 = 0
√(x-1) >= 0 and
(xy-3)^2 >= 0
so √(x-1) = 0 and (xy-3)^2 = 0 for √(x-1) + (xy-3)^2 = 0
(i.e.) x = 1 and y = 3

1/xy + 1/(x+1)(y+1) + 1/(x+2)(y+2) + ... + 1/(x+22)(y+22)
= 1/(1*3) + 1/(2*4) + 1/(3*5) + ... + 1/(23*25)
= (1/2)[1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + ... + (1/21 - 1/23) +
(1/22 - 1/24) + (1/23 - 1/25) ]
= (1/2)(1 + 1/2 - 1/24 - 1/25)
= (1/2)(1 + 1/2 - 49/600)
= (1/2)(851/600)
= 851/1200