Question

3а+2/а²-2а+1 - 6/а²-1 - 3а-2/а²+2а+1
Как решить?

Answer 1

$\frac{3a+2}{ a^{2} -2a+1} - \frac{6}{a^{2} -1} - \frac{3a-2}{a^{2}+2a+1} = \frac{3a+2}{(a-1)(a-1)}- \frac{6}{(a-1)(a+1)} - \frac{3a-2}{(a+1)(a+1)} = \\ \\ \frac{(3a+2)(a+1)(a+1)-6(a-1)(a+1)-(3a-2)(a-1)(a-1)}{(a-1)(a+1)(a-1)(a+1)} = \\ \\ \frac{(3a+2)( a^{2} +2a+1)-6a^{2} +6-(3a-2)( a^{2}-2a+1) }{(a-1)(a+1)(a-1)(a+1)} = \\ \\ \frac{3a^{3} +6a^{2}+3a+2a^{2}+4a+2-6 a^{2} +6-3a^{3}+6a^{2} -3a+2 a^{2} -4a+2}{(a-1)(a+1)(a-1)(a+1)}= \\ \\ [tex]\frac{10a^{2}+10 }{(a-1)(a+1)(a-1)(a+1)}= \frac{10(a^{2} +1)}{(a^{2} -1)(a^{2} -1)}$