Question

доказать тождество (a+1)(a-1)-(a+3)²=6a-10​

Answer 1

Ответ:

$(a+1)(a-1) = a^{2} - 1\\\\(a+3)^{2} = a^{2} + 6a + 9\\ \\ (a+1)(a-1)-(a+3)^{2} = a^{2} - 1 - a^{2} - 6a - 9 = -6a - 10$