Question

Помогите, пожалуйста, с алгеброй!!!
Только 36 номер!!!

Answer 1

36.

#1.

$\frac{c + 6}{ {c}^{2} - 4c + 4} \div \frac{ {c}^{2} - 36}{16c - 32} - \frac{4}{c - 6} = \frac{4}{2 - c}$

1)

$\frac{c + 6}{ {c}^{2} - 4c + 4} \div \frac{ {c}^{2} - 36}{16c - 32} = \frac{c + 6}{ {(c - 2)}^{2} } \div \frac{ {c}^{2} - {6}^{2} }{16(c - 2)} = \frac{c + 6}{ (c - 2)(c - 2) } \div \frac{(c -6)(c + 6) }{16(c - 2)} = \frac{c + 6}{(c - 2)(c - 2)} \times \frac{16(c - 2)}{(c - 6)(c + 6)} = \frac{16}{(c - 2)(c - 6)}$

2)

$\frac{16}{(c - 2)(c - 6)} - \frac{4}{(c - 6)} = \frac{16}{(c - 2)(c - 6)} - \frac{4 \times (c - 2)}{(c - 6) \times (c - 2)} = \frac{16 - 4c + 8}{(c - 6)(c - 2)} = \frac{24 - 4c}{(c - 6)(c - 2)} = \frac{ - 4(c - 6)}{(c - 6)(c - 2)} = \frac{ - 4}{c - 2} = - \frac{4}{c - 2} = \frac{4}{2 - c}$

Итого,

$\frac{4}{2 - c} = \frac{4}{2 - c}$

#2.

$( \frac{1}{ {(x - 7)}^{2} } + \frac{2}{ {x}^{2} - 49} + \frac{1}{ {(x + 7)}^{2} } ) \div \frac{16 {x}^{4} }{ {( {x}^{2} - 49)}^{2} } = \frac{1}{4 {x}^{2} }$

1)

$\frac{1}{ {(x - 7)}^{2} } = \frac{1}{(x - 7)(x - 7)}$

,

$\frac{2}{ {x}^{2} - 49} = \frac{2}{ {x}^{2} - {7}^{2} } = \frac{2}{(x - 7)(x + 7)}$

,

$\frac{1}{ {(x + 7)}^{2} } = \frac{1}{(x + 7)(x + 7)}$

. Общий знаменатель у дробей —

$(x - 7)(x - 7)(x + 7)(x + 7)$

.

Итак, в скобках:

$\frac{(x + 7)(x + 7) + 2(x - 7)(x + 7) + (x - 7)(x - 7)}{(x - 7)(x - 7)(x + 7)(x + 7)} = \frac{ {x}^{2} + 14x + 49 + 2 {x}^{2} - 98 + {x}^{2} - 14x + 49}{(x - 7)(x - 7)(x + 7)(x + 7)} = \frac{4 {x}^{2} }{(x - 7)(x - 7)(x + 7)(x + 7)} = \frac{4 {x}^{2} }{(x - 7)(x + 7)(x - 7)(x + 7)} = \frac{4 {x}^{2} }{( {x}^{2} - {7}^{2} )( {x}^{2} - {7}^{2} ) } = \frac{4 {x}^{2} }{ {( {x}^{2} - 49)}^{2} }$

2)

$\frac{4 {x}^{2} }{ {( {x}^{2} - 49)}^{2} } \div \frac{16 {x}^{4} }{ {( {x}^{2} - 49)}^{2} } = \frac{4 {x}^{2} }{ {( {x}^{2} - 49 )}^{2} } \times \frac{ {( {x}^{2} - 49 )}^{2} }{16 {x}^{4} } = \frac{1}{4 {x}^{2} }$

Итого,

$\frac{1}{4 {x}^{2} } = \frac{1}{4 {x}^{2} }$