`d, {( (5-x)^2 - 1/[y-1] = 4),( (5-x)^2 + 3/[1-y]  = 2):} (y ne 1)`

`<=> {( -4/[y-1]  = 2),( (5-x)^2 = 4+  1/[y-1]):}`

`<=> {( 1/[y-1] = -1/2),( (5-x)^2 = 4 - 1/2 = 7/2):}`

`<=> {( y -1= -2),( (x-5)^2 = 7/2):}`

`<=> {( y = -1),( x = [10 +- sqrt14]/2):} ( N)`

Vậy `(x;y) = ( [10 +-sqrt14]/2 ; -1)`