Ответ:
Упростить выражение . Применяем формулы сокращённого умножения .
[tex]\bf \displaystyle \frac{a+11}{a+9}-\Big(\frac{a+5}{a^2-81}+\frac{a+7}{a^2-18a+81}\Big)\, :\, \Big(\frac{a+3}{a-9}\Big)^2=\\\\\\=\frac{a+11}{a+9}-\Big(\frac{a+5}{(a-9)(a+9)}+\frac{a+7}{(a-9)^2}\Big)\, :\, \frac{(a+3)^2}{(a-9)^2}=\\\\\\=\frac{a+11}{a+9}-\Big(\frac{(a+5)(a-9)+(a+7)(a+9)}{(a-9)^2(a+9)}\Big)\cdot\, \frac{(a-9)^2}{(a+3)^2}=\\\\\\=\frac{a+11}{a+9}-\frac{(\a^2-4a-45+a^2+16a+63}{(a-9)^2(a+9)}\cdot\, \frac{(a-9)^2}{(a+3)^2}=[/tex]
[tex]\bf \displaystyle =\frac{a+11}{a+9}-\frac{2a^2+12a+18}{a+9}\cdot\, \frac{1}{(a+3)^2}=\\\\\\=\frac{a+11}{a+9}-\frac{2(a^2+6a+9)}{a+9}\cdot\, \frac{1}{(a+3)^2}=\\\\\\=\frac{a+11}{a+9}-\frac{2(a+3)^2}{a+9}\cdot\, \frac{1}{(a+3)^2}=\frac{a+11}{a+9}-\frac{2}{a+9}=\frac{a+11-2}{a+9}=\\\\\\=\frac{a+9}{a+9}=1[/tex]
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Ответ:
Упростить выражение . Применяем формулы сокращённого умножения .
[tex]\bf \displaystyle \frac{a+11}{a+9}-\Big(\frac{a+5}{a^2-81}+\frac{a+7}{a^2-18a+81}\Big)\, :\, \Big(\frac{a+3}{a-9}\Big)^2=\\\\\\=\frac{a+11}{a+9}-\Big(\frac{a+5}{(a-9)(a+9)}+\frac{a+7}{(a-9)^2}\Big)\, :\, \frac{(a+3)^2}{(a-9)^2}=\\\\\\=\frac{a+11}{a+9}-\Big(\frac{(a+5)(a-9)+(a+7)(a+9)}{(a-9)^2(a+9)}\Big)\cdot\, \frac{(a-9)^2}{(a+3)^2}=\\\\\\=\frac{a+11}{a+9}-\frac{(\a^2-4a-45+a^2+16a+63}{(a-9)^2(a+9)}\cdot\, \frac{(a-9)^2}{(a+3)^2}=[/tex]
[tex]\bf \displaystyle =\frac{a+11}{a+9}-\frac{2a^2+12a+18}{a+9}\cdot\, \frac{1}{(a+3)^2}=\\\\\\=\frac{a+11}{a+9}-\frac{2(a^2+6a+9)}{a+9}\cdot\, \frac{1}{(a+3)^2}=\\\\\\=\frac{a+11}{a+9}-\frac{2(a+3)^2}{a+9}\cdot\, \frac{1}{(a+3)^2}=\frac{a+11}{a+9}-\frac{2}{a+9}=\frac{a+11-2}{a+9}=\\\\\\=\frac{a+9}{a+9}=1[/tex]