Question

Представьте в виде рациональной дроби
(a/b - b/a) : a-b/3ab = 3a+3b
(x-3/x+3 - x^2 + 27/x^2 - 9) * x^2 - 6x + 9/6 = 3-x
(3a + 3/a^2-1 - a/a-1) : a-3/1-a = 1

Answer 1

$(\frac{a}{b} -\frac{b}{a}) : \frac{a-b}{3ab} =$

$=(\frac{a^2}{ab} -\frac{b^2}{ab})\cdot \frac{3ab}{a-b} =$

$=(\frac{a^2 -b^2}{ab})\cdot \frac{3ab}{a-b} =$

$=\frac{(a-b)(a+b)}{1}\cdot \frac{3}{a-b} =$

$=3(a+b)=3a+3b$

--------------------

$(\frac{x-3}{x+3} - \frac{x^2 + 27}{x^2 - 9})\cdot\frac{ x^2 - 6x + 9}{6} =$

$=(\frac{x-3}{x+3} - \frac{x^2 + 27}{(x-3)(x+3)})\cdot\frac{(x-3)^2}{6} =$

$=(\frac{(x-3)^2}{(x-3)(x+3)} - \frac{x^2 + 27}{(x-3)(x+3)})\cdot\frac{(x-3)^2}{6} =$

$=\frac{(x-3)^2 - (x^2 + 27)}{(x-3)(x+3)})\cdot\frac{(x-3)^2}{6} =$

$=\frac{x^2-6x+9-x^2-27}{x+3})\cdot\frac{x-3}{6} =$

$=\frac{-6x-18}{x+3})\cdot\frac{x-3}{6} =$

$=\frac{-6(x+3)}{x+3})\cdot\frac{x-3}{6} =$

$=-(x-3) =-x+3=3-x$

--------------------

$(\frac{3a + 3}{a^2-1} - \frac{a}{a-1}) :\frac{ a-3}{1-a} =$

$=(\frac{3(a+1)}{(a-1)(a+1)} - \frac{a}{a-1}) \cdot\frac{1-a}{ a-3} =$

$=(\frac{3}{a-1} - \frac{a}{a-1}) \cdot\frac{1-a}{ a-3} =$

$=\frac{3 -a}{a-1}) \cdot\frac{1-a}{ a-3} =$

$=\frac{-(a-3)}{-(1-a)}) \cdot\frac{1-a}{ a-3} =$

$=\frac{-1}{-1}=1$