Question

Пожалуйсто помогите со 2, 3 ,4, 5,6,7 позязяяяяяя 25 б

Answer 1

$2)\; \; \frac{x+4}{3x-1}\; \; \to \; \; 3x-1\ne 0\; ,\; \; x\ne \frac{1}{3}\\\\3)\; \; \frac{2t-1}{3t}-\frac{1-t}{3t}=\frac{2t-1-(1-t)}{3t}=\frac{2t-1-1+t}{3t}=\frac{3t-2}{3t}\\\\4)\; \; \frac{a^2-6a+9}{a^2-9}=\frac{(a-3)^2}{(a-3)(a+3)}=\frac{a-3}{a+3}\\\\5)\; \; \frac{b-5}{b-3}\cdot \frac{b^2-9}{5b-b^2}=\frac{b-5}{b-3}\cdot \frac{(b-3)(b+3)}{b(5-b)}=-\frac{b+3}{b}$

$6)\; \; \frac{2x^2}{x-5}-2x=\frac{2x^2-2x(x-5)}{x-5}=\frac{2x^2-2x^2+10x}{x-5}=\frac{10x}{x-5}\\\\7)\; \; \frac{y^2-10y+25}{5y-1}:(5-y)^2=\frac{(5-y)^2}{(5y-1)\cdot (5-y)^2}=\frac{1}{5y-1}$