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a, x/ x-2 = x-6 / x-4
=> x(x-4) /(x-2)(x-4) = (x-6)(x-2) / (x-4)(x-2)
=> x^2 - 4x = x^2 - 2x - 6x + 12
=> x^2 - x^2 -4x + 2x + 6x = 12
=> 4x = 12
=> x=3
Vậy ...
b, 3x-5/x-1 - 2x-5/x-2 = 1
=> (3x-5)(x-2) / (x-1)(x-2) - (2x-5)(x-1) / (x-1)(x-2) = (x-1)(x-2) / (x-1)(x-2)
=> (3x-5)(x-2) - (2x-5)(x-1) = (x-1)(x-2)
=> 3x^2 - 6x - 5x +10 - ( 2x^2 - 2x -5x + 5) = x^2 - 2x - x + 2
=> 3x^2 - 6x - 5x +10 - 2x^2 + 2x + 5x - 5 = x^2 - 2x - x + 2
=> 3x^2 - 2x^2 - x^2 -6x - 5x + 2x + 5x + 2x + x = -10 + 5 +2
=> -x = -3
=> x=3
Vậy...
c, x+1/x-1 - 4/x+1 + x^2 - 3 / 1-x^2 = 0
x+1/x-1 - 4/x+1 - x^2 - 3 / x^2 - 1 = 0
(x+1)^2/(x-1)(x+1) - 4(x-1)/(x+1)(x-1) - x^2 - 3 / (x-1)(x+1)= 0
x^2 + 2x + 1 - 4x + 4 + x^2 - 3= 0
x^2 - x^2 + 2x-4x = -1-4+3
-2x = -2
x=1
Vậy....
Chấm nha