Question

Ответьте пожалуйста даю 20балов умоляю

Answer 1

Объяснение:

№1)

$\bigg( \frac{a + b}{a - b} - \frac{a - b}{a + b}\bigg) : \frac{ab}{a {}^{2} - {b}^{2} } = \\ = \bigg( \frac{a + b}{a - b} - \frac{a - b}{a + b}\bigg) \times \frac{a {}^{2} - {b}^{2} } {ab}= \\ = \bigg( \frac{(a + b)(a + b)}{(a - b)(a + b)} - \frac{(a - b)(a - b)}{(a + b)(a - b)}\bigg) \times \frac{a {}^{2} - {b}^{2} } {ab}= \\ \bigg( \frac{(a + b) {}^{2} }{(a - b)(a + b)} - \frac{(a - b) {}^{2} }{(a - b)(a + b) }\bigg) \times \frac{(a - b)(a + b)} {ab}= \\ \bigg( \frac{(a + b)^{2} - (a - b) {}^{2} }{(a - b)(a + b) }\bigg) \times \frac{(a - b)(a + b)} {ab}= \\ = \frac{(a + b)^{2} - (a - b) {}^{2} }{(a - b)(a + b) } \times \frac{(a - b)(a + b)} {ab}= \\ = \frac{(a^{2} + 2ab + b^{2} ) - (a^{2} - 2ab + b^{2} ) }{(a - b)(a + b) } \times \frac{(a - b)(a + b)} {ab}= \\ = \frac{ \cancel{a^{2}} + 2ab + \cancel{b^{2}} - \cancel{a^{2}} + 2ab - \cancel{b^{2}} }{(a - b)(a + b) } \times \frac{(a - b)(a + b)} {ab}= \\ = \frac{4{ \cdot}\cancel{ab \: }}{{\cancel{(a - b)}\cancel{(a + b) }} } \times \frac{{\cancel{(a - b)}\cancel{(a + b)} } }{\cancel{ab}} = 4$

№2)

$\frac{{a}^{2} + 9}{ {a}^{2} - 9} - \frac{a}{a - 3} = \\ = \frac{{a}^{2} + 9}{(a - 3)(a + 3)} - \frac{a(a + 3)}{(a - 3)(a + 3)} = \\ = \frac{{a}^{2} + 9 - a(a + 3)}{(a - 3)(a + 3)} = \\ = \frac{{a}^{2} + 9 - a^{2} - 3a}{(a - 3)(a + 3)} = \frac{ 9 - 3a}{(a - 3)(a + 3)} =\\ = \frac{ 3(3 - a)}{(a - 3)(a + 3)} = \frac{ - 3 \cancel{(a - 3)}}{ \cancel{(a - 3)}(a + 3)} =\\ = - \frac{3}{a + 3}$

$\frac{c + 1}{3c} : \frac{ {c}^{2} - 1 }{6 {c}^{2} } = \frac{c + 1}{3c} \times \frac{6 {c}^{2} } { {c}^{2} - 1 } =\\ = \frac{ \cancel{ \: c + 1 \: }}{ {3 \: } \: \cancel{c \: }} \times \frac{ {6 \:} {c}^{ \: \cancel{2 \: }} } { ({c}- 1 ) \cancel{(c + 1)}} = \frac{6c}{3(c - 1)} \\ = \frac{2c}{c - 1}$

№3)

1)

$(10^3)^2 \cdot 10^8= (10^{3\cdot2 }\cdot 10^8=\\=10^{6+8}=10^{14}$

2),3),4) - см. на фото