Question

Помогите решить 5 и 6 задание​

Answer 1

Объяснение:

5)

$\frac{72 {a}^{3}b }{c} \div (27 {a}^{2} b) = \frac{72 {a}^{3}b }{c} \times \frac{1}{27 {a}^{2}b} = \frac{72 {a}^{3} }{c} \times \frac{1}{ {27a}^{2} } = \frac{72 {a}^{3} }{ {27a}^{2} c} = \frac{8a}{3c}$

6)

$\frac{ {4a}^{2} - 1 }{ {a}^{2} - 9 } \div \frac{10a + 5}{a + 3} = \frac{(2a - 1) \times (2a + 1)}{(a - 3) \times (a + 3)} \times \frac{a + 3}{10a + 5} = \frac{(2a - 1) \times (2a + 1)}{(a - 3)} \times \frac{1}{5(2a + 1)} = \frac{(2a - 1)}{(a - 3)} \times \frac{1}{5} = \frac{(2a - 1)}{5(a - 3)} = \frac{2a - 1}{5a - 15}$

Answer 2

Ответ:

$5) \frac{72a^{3}b }{c}/(27a^{2}b)= \frac{72a^{3}b }{c}*\frac{1}{27a^{2}b} =\frac{72a^{3} b}{27a^{2}bc}=\boxed{\frac{8a}{3c}}$

$6)\frac{4a^{2}-1 }{a^{2}-9 } / \frac{10a+5}{a+3}= \frac{(2a-1)(2a+1)}{(a-3)(a+3)}*\frac{a+3}{10a+5}=\frac{(2a-1)(2a+1)}{a-3}*\frac{1}{5(2a+1)}=\frac{2a-1}{a-3}*\frac{1}{5}=\frac{1(2a-1)}{5(a-3)}=\boxed{\frac{2a-1}{5a-15}}$

Объяснение: