Question

помогите пожайлусто мне не справится ни как

Answer 1

Решения и ответы прикреплены.

Answer 2

a)$( \frac{a}{3}-\frac{a}{4})\frac{2}{a^2}=\frac{4a-3a}{3*4} \frac{2}{a^2}=\frac{a*2}{12*a^2}=\frac{1}{6a}$
б)$\frac{a-1}{3a}:(1- \frac{1}{a})=\frac{a-1}{3a}:(\frac{a}{a}- \frac{1}{a} )=\frac{a-1}{3a}:(\frac{a-1}{a} )=$
$=\frac{a-1}{3a}*\frac{a}{a-1} = \frac{a(a-1)}{3a(a-1)}= \frac{1}{3}$
в)$(1+ \frac{y}{x} ):(1- \frac{y}{x} )=( \frac{x}{x} + \frac{y}{x} ):( \frac{x}{x} - \frac{y}{x} )= \frac{x+y}{x}: \frac{x-y}{x}= \frac{x+y}{x}* \frac{x}{x-y}=$
$=\frac{x(x+y)}{x(x-y)}= \frac{x+y}{x-y}$

a)$\frac{a}{1-b}+ \frac{a-ab}{b^2-1}* \frac{b+1}{a}= \frac{a}{1-b}+ \frac{a(1-b)}{(b-1)(b+1)}* \frac{b+1}{a} =$
$= \frac{a}{1-b}- \frac{a(1-b)(b+1)}{a(1-b)(b+1)} =\frac{a}{1-b}-1= \frac{a}{1-b}- \frac{1-b}{1-b}=\frac{a-1+b}{1-b}=\frac{a+b-1}{1-b}$
б) $\frac{a}{a+2}: \frac{a^2+3a}{a^2-4}- \frac{a}{a+3}=\frac{a}{a+2}*\frac{a^2-4}{a^2+3a}- \frac{a}{a+3}=$
=$\frac{a}{a+2}*\frac{(a-2)(a+2)}{a(a+3)}- \frac{a}{a+3}=\frac{a(a-2)(a+2)*}{a(a+3)(a+2)}- \frac{a}{a+3}=$
$= \frac{a-2}{a+3} - \frac{a}{a+3}=\frac{a12-a}{a+3}=-\frac{2}{a+3}$

$1-\frac{1}{ 1-\frac{1}{x-3} }= 1-\frac{1}{ \frac{x-3}{x-3}-\frac{1}{x-3} }=1-\frac{1}{ \frac{x-3-1}{x-3}}=1-\frac{x-3}{ x-4}=\frac{x-4}{ x-4}-\frac{x-3}{ x-4}=$
=$\frac{x-4-x+3}{ x-4}=\frac{-1}{ x-4}=\frac{1}{ 4-x}$