Question

plz answer this question .it's very important to me​

Answer 1

Answer:

$= > \frac{1}{x - 1}+\frac{1}{x - 2} = \frac{1}{x - 3} \\ \\ = > \frac{x - 2 + x - 1}{(x - 1)(x - 2)} = \frac{1}{x - 3} \\ \\ = > \frac{2x - 3}{(x - 1)(x - 2)} = \frac{1}{x - 3} \\ \\ = > (2x - 3)(x - 3) = (x - 1)(x - 2) \\ \\ = > 2 {x}^{2} + 9 - 9x = {x}^{2} + 2 - 3x \\ \\ = > {x}^{2} - 6x + 9 = 0 \\ \\ \\ = > d = {b}^{2} - 4ac \\ \: \: \: \: \: \: \: \: \: \: \: \: = 36 - 36 \\ \: \: \: \: \: \: \: \: \: \: \: \: \ = 0 \\ \\ since \: \: \: d = 0 \\ \: the \: equation \: has \: one \: real \: root$