Question

Помогите пожалуйста с заданиями 1, 2, 3.
Даю 80 балов

Answer 1

1.

$\frac{a}{b} \\ \\ a \in R; \ \ b\neq 0$

$\frac{3}{a-1} \\ \\ a-1\neq 0 \\ a\neq 1$

$\frac{a}{b+2} \\ \\ a \in R; \ \ \ b+2\neq 0 \\ \\ b\neq -2$

2.

$4a+\frac{1-4a^2}{a}=\frac{4a\cdot a+1-4a^2}{a}=\frac{4a^2+1-4a^2}{a}=\frac{1}{a} \\ \\ \frac{a+b}{a-b}-\frac{a-b}{a+b}=\frac{(a+b)\cdot (a+b)-(a-b)\cdot (a-b)}{(a-b)\cdot (a+b)}=\frac{a^2+2ab+b^2 - (a^2-2ab+b^2)}{a^2-b^2}=\frac{a^2+2ab+b^2-a^2+2ab-b^2}{a^2-b^2}=\\ \\ =\frac{4ab}{a^2-b^2}$

$\frac{2a-4}{3b}\cdot \frac{6b}{a-2 }=\frac{2\cdot( a-2)}{1}\cdot \frac{2}{a-2 }=2\cdot 2=4 \\ \\ \frac{a^2-b^2}{b^2} : \frac{a+b}{b}=\frac{a^2-b^2}{b^2} \cdot \frac{b}{a+b}=\frac{(a-b)\cdot (a+b)}{b}\cdot\frac{1}{a+b}=\frac{a-b}{b}=\frac{a}{b}-1$

3.

$\frac{1+2x}{x-3}-\frac{x^2+3x}{5}\cdot \frac{10}{x^2-9}=\frac{1+2x}{x-3}-\frac{x\cdot (x+3)}{5}\cdot \frac{10}{(x-3)\cdot (x+3)}=\frac{1+2x}{x-3}-\frac{x}{1}\cdot \frac{2}{x-3}= \\ \\ = \frac{1+2x}{x-3}-\frac{2x}{x-3}=\frac{1+2x-2x}{x-3}=\frac{1}{x-3} \\ \\ x=2\frac{2}{3} \\ \\ \frac{1}{2\frac{2}{3}-3 }=\frac{1}{\frac{8}{3}-3}=\frac{1}{-\frac{1}{3}}=-3$