6)
[tex]1)\\\\\frac{9}{5+b}-\frac{4-b}{5+b}=\frac{9-(4-b)}{5+b}=\frac{9-4+b}{5+b}=\frac{5+b}{5+b}=1\\\\2)\\\frac{a+3}{a-6}+\frac{a-15}{a-6}=\frac{a+3+a-15}{a-6}=\frac{2a-12}{a-6}=\frac{2(a-6)}{a-6}=2[/tex]
7)
[tex]1)\\\\\frac{10p^2-n}{2p}-5p=\frac{10p^2-n}{2p}-\frac{5p\cdot 2p}{2p}=\frac{10p^2-n}{2p}-\frac{10p^2}{2p}=\\\\\frac{10p^2-n-10p^2}{2p}=\frac{-n}{2p}=-\frac{n}{2p}\\\\2)\\\\\frac{2a}{b^2-4a^2}+\frac{1}{2a-b}=-\frac{2a}{(2a)^2-b^2}+\frac{1}{2a-b}=\\\\-\frac{2a}{(2a-b)(2a+b)}+\frac{2a+b}{(2a-b)(2a+b)}=\frac{-2a+2a+b}{(2a-b)(2a+b)}=\frac{b}{(2a-b)(2a+b)}[/tex]
[tex]3)\\\\\frac{25}{x^2-5x}-\frac{x}{x-5}+\frac{x+5}{x}=\frac{25}{x(x-5)}-\frac{x^2}{x(x-5)}+\frac{(x+5)(x-5)}{x(x-5)}=\\\\\frac{25-x^2+(x+5)(x-5)}{x(x-5)}=\frac{25-x^2+x^2-25}{x(x-5)}=\frac{0}{x(x-5)}=0[/tex]
8)
ОДЗ:
[tex]x\not=-4\\\\y=\frac{x^2+4x}{8+2x}=\frac{x(x+4)}{2(4+x)}=\frac{1}{2}x[/tex]
на фото
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
6)
[tex]1)\\\\\frac{9}{5+b}-\frac{4-b}{5+b}=\frac{9-(4-b)}{5+b}=\frac{9-4+b}{5+b}=\frac{5+b}{5+b}=1\\\\2)\\\frac{a+3}{a-6}+\frac{a-15}{a-6}=\frac{a+3+a-15}{a-6}=\frac{2a-12}{a-6}=\frac{2(a-6)}{a-6}=2[/tex]
7)
[tex]1)\\\\\frac{10p^2-n}{2p}-5p=\frac{10p^2-n}{2p}-\frac{5p\cdot 2p}{2p}=\frac{10p^2-n}{2p}-\frac{10p^2}{2p}=\\\\\frac{10p^2-n-10p^2}{2p}=\frac{-n}{2p}=-\frac{n}{2p}\\\\2)\\\\\frac{2a}{b^2-4a^2}+\frac{1}{2a-b}=-\frac{2a}{(2a)^2-b^2}+\frac{1}{2a-b}=\\\\-\frac{2a}{(2a-b)(2a+b)}+\frac{2a+b}{(2a-b)(2a+b)}=\frac{-2a+2a+b}{(2a-b)(2a+b)}=\frac{b}{(2a-b)(2a+b)}[/tex]
[tex]3)\\\\\frac{25}{x^2-5x}-\frac{x}{x-5}+\frac{x+5}{x}=\frac{25}{x(x-5)}-\frac{x^2}{x(x-5)}+\frac{(x+5)(x-5)}{x(x-5)}=\\\\\frac{25-x^2+(x+5)(x-5)}{x(x-5)}=\frac{25-x^2+x^2-25}{x(x-5)}=\frac{0}{x(x-5)}=0[/tex]
8)
ОДЗ:
[tex]x\not=-4\\\\y=\frac{x^2+4x}{8+2x}=\frac{x(x+4)}{2(4+x)}=\frac{1}{2}x[/tex]
на фото