Question

Помогите пожалуйста с примерами

Answer 1

Ответ:

1) 0

2) 5

3) 2x

4) -1/a

5) 0

Объяснение:

$1) \frac{9a^{2}-4 }{2-3a} - \frac{6a^{2}-5a-6}{3-2a} = \\\\\\\ = \frac{(3a-2)(3a+2)}{2-3a} - \frac{(2a-3)(3a+2)}{3-2a} = \\\\= -(3a+2) + (3a+2) = 0$

$2) 5* \frac{(a^{2}+5a+6)(a^{2}-2a+4)}{(a+3)(a^{3}+8)} = \\\\\\= 5* \frac{(a+2)(a+3)(a^{2}-2a+4}{(a+3)(a+2)(a^{2}-2a+4)} = 5$

$3) \frac{4x^{2}-5x+1}{4x-1} - \frac{x^{2}-1}{1-x} = \\\\\\= \frac{(4x-1)(x-1)}{4x-1} - \frac{(x-1)(x+1)}{1-x} = (x-1) + (x+1) = 2x$

$4) (\frac{1}{2-6a} + \frac{1}{27a^{3}-1} : \frac{1+3a}{1+3a+9a^{2}} ) * \frac{2+6a}{a} = \\\\\\= ( \frac{1}{2(1-3a)} + \frac{1}{(3a-1)(9a^{2}+3a+1)} : \frac{1+3a}{1+3a+9a^{2}} ) * \frac{2(1+3a)}{a} = \\\\\\= ( \frac{1}{2(1-3a)} + \frac{1}{(3a-1)(3a+1)} ) * \frac{2(1+3a)}{a} = \\\\\\= \frac{3a+1-2}{2(1-3a)(1+3a)} * \frac{2(1+3a)}{a} = - \frac{1}{a}$

$5) \frac{2p^{3}}{p^{3}+q^{3}} * \frac{p+q}{p} - \frac{2p^{2}}{p^{2}-pq+q^{2}} = \\\\\\= \frac{2p^{2}}{p^{2}-pq+q^{2}} - \frac{2p^{2}}{p^{2}-pq+q^{2}} = 0$