Berúc do úvahy danú eqution so zmenou # 4 (1 + y) x ^ 2-4xy- (1-y) # # => 4 (1 + y) x ^ 2-2 (1 + y) x + 2 (1-y) X- (1-y) # # => 2 (1 + y) x (2 x-1) + (1-y) (2 x-1) # # => (2x-1), (2 (1 + y) x + (1-y)) = 0 # z toho dôvodu # X = 1/2 # kontrola # 4 (1 + y) x ^ 2-4xy- (1-y) # # = 4 (1 + y) (1/2) ^ 2-4 (1/2) y- (1-y) # # = 1 + y-2y-1 + y = 0 #
