\[ \frac{1 \cdot (x-1) + 1 \cdot (2x-3)}{(2x-3)(x-1)} = 2 \]
\[ \frac{x-1 + 2x-3}{(2x-3)(x-1)} = 2 \]
\[ \frac{3x-4}{(2x-3)(x-1)} = 2 \]
\[ (2x-3)(x-1) = 2x^2 - 2x - 3x + 3 = 2x^2 - 5x + 3 \]
\[ \frac{3x-4}{2x^2 - 5x + 3} = 2 \]
\[ 3x-4 = 2(2x^2 - 5x + 3) \]
\[ 3x-4 = 4x^2 - 10x + 6 \]
\[ 4x^2 - 10x - 3x + 6 + 4 = 0 \]
\[ 4x^2 - 13x + 10 = 0 \]
\[ D = b^2 - 4ac = (-13)^2 - 4 \cdot 4 \cdot 10 = 169 - 160 = 9 \]
\[ \sqrt{D} = 3 \]
\[ x_1 = \frac{-b - \sqrt{D}}{2a} = \frac{13 - 3}{2 \cdot 4} = \frac{10}{8} = \frac{5}{4} \]
\[ x_2 = \frac{-b + \sqrt{D}}{2a} = \frac{13 + 3}{2 \cdot 4} = \frac{16}{8} = 2 \]
При x = 5/4:
2x - 3 = 2(5/4) - 3 = 5/2 - 3 = 5/2 - 6/2 = -1/2
eq 0
x - 1 = 5/4 - 1 = 1/4
eq 0
При x = 2:
2x - 3 = 2(2) - 3 = 4 - 3 = 1
eq 0
x - 1 = 2 - 1 = 1
eq 0
Ответ: 5/4