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Question

(x-12+y).(x-12+y)+(y+4-x).(y+4-x)=0

=>(x;y)=(...;...)

Answer 1

(x-12+y).(x-12+y)+(y+4-x).(y+4-x) = 0

=> (x-12+y)^2 + (y+4-x)^2 = 0

=> x - 12 +y = 0

     y +4 - x = 0

=> x+y = 12

      y -x = -4

=> x = [12-(-4)]:2 = 8

     y = 12-8 = 4

vậy x = 8, y = 4

Comments

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