Question

Спростити вирази 1
2.
Очень нужно

$(1 - \frac{1}{12y} ) \div \frac{1}{3y}$
$( \frac{c + n}{n - c} - 1) \div \frac{4}{c - n}$

Answer 1

$(1-\frac{1}{12y} ): \frac{1}{3y} =\frac{12y-1}{4}$

$(\frac{c+n}{n-c} -1):\frac{4}{c-n} =-\frac{c}{2}$

Объяснение:

1 .

$(1-\frac{1}{12y} ): \frac{1}{3y} =(\frac{12y}{12y}-\frac{1}{12y} ): \frac{1}{3y}=\frac{12y-1}{12y}* \frac{3y}{1}=\frac{12y-1}{4}$

2.

Поскольку

$\\\frac{c+n}{n-c}=\frac{c+n}{-(-n+c)}=-\frac{c+n}{c-n}=\frac{-(c+n)}{c-n}=\frac{-c-n}{c-n}\\$

то

$(\frac{c+n}{n-c} -1):\frac{4}{c-n} =(\frac{-c-n}{c-n}-\frac{c-n}{c-n}):\frac{4}{c-n} =\\=\frac{-c-n-(c-n)}{c-n}:\frac{4}{c-n} =\frac{-c-n-c+n}{c-n}:\frac{4}{c-n} =\\=\frac{-2c}{c-n}*\frac{c-n}{4}=\frac{-c}{2}=-\frac{c}{2}$