Simplify: \( \rm \frac{1 - 2a}{4a^{2} - 1} - \frac{a - 1}{2a^{2} - 3a + 1} - \frac{1}{1 - a} \).
Atith Adhikari
·
2 years ago
Solution
Given,
\( \rm \frac{1 - 2a}{4a^{2} - 1} - \frac{a - 1}{2a^{2} - 3a + 1} - \frac{1}{1 - a} \)
\( \rm = \frac{1 - 2a}{ (2a)^{2} - 1^{2}} - \frac{a -1 }{2a^{2} - (2 + 1)a + 1} - \frac{1}{1 - a} \)
\( \rm = \frac{1 - 2a}{(2a + 1)(2a - 1)} - \frac{a - 1}{2a^{2} - 2a - a + 1} - \frac{1}{1 - a} \)
\( \rm = \frac{1 - 2a}{(2a + 1)(2a - 1)} - \frac{a - 1}{2a (a - 1) - 1(a - 1)} - \frac{1}{1 - a} \)
\( \rm = \frac{1 - 2a}{(2a + 1)(2a - 1)} - \frac{a - 1}{(2a -1)(a - 1)} - \frac{1}{1 - a} \)
\( \rm = \frac{ (1 - 2a)(a - 1) - (a-1)(2a + 1)}{(2a - 1)(2a + 1)(a - 1)} - \frac{1}{1 - a} \)
\( \rm = \frac{ (a - 1 - 2a^{2} + 2a) - (2a^{2} + a - 2a -1 )}{(2a -1)(2a + 1)(a - 1)} - \frac{1}{-(1 - a)} \)
\( \rm = \frac{ 3a - 1 - 2a^{2} - 2a^{2} + a + 1}{(2a - 1)(2a + 1)(a - 1)} + \frac{1}{a - 1} \)
\( \rm = \frac{4a - 4a^{2}}{(2a - 1)(2a + 1)(a - 1)} + \frac{1}{a - 1} \)
\( \rm = \frac{4a - 4a^{2}}{(2a - 1)(2a + 1)(a - 1)} + \frac{(2a + 1)(2a - 1)}{(2a - 1)(2a + 1)(a - 1)} \)
\( \rm = \frac{4a - 4a^{2} + (4a^{2} - 1)}{(2a - 1)(2a + 1)(a - 1)} \)
\( \rm = \frac{4a - 4a^{2} + 4a^{2} + 1}{(2a - 1)(2a + 1)(a - 1)} \)
\( \rm = \frac{4a - 1}{(2a - 1)(2a + 1)(a - 1)} \)