[tex]\left (\left (\cfrac{3a}{a^2-b^2}\cdot \cfrac{a^2+b^2+ab}{a+b}-\cfrac{3}{b-a} \right ):\cfrac{2a+b}{a^2+2ab+b^2} \right ):\cfrac{3}{a+b}[/tex]
Первое преобразование
[tex]\cfrac{3a}{a^2-b^2}\cdot \cfrac{a^2+b^2+ab}{a+b}-\cfrac{3}{b-a}=\cfrac{3a}{(a-b)(a+b)}\cdot \cfrac{a^2+2ab+b^2-ab}{a+b}+\cfrac{3}{a-b}=\\=\cfrac{3a}{(a-b)(a+b)}\cdot \cfrac{(a+b)^2-ab}{a+b}+\cfrac{3}{a-b}=\\=\cfrac{3a}{(a-b)(a+b)}\left ( a+b-\cfrac{ab}{a+b} \right )+\cfrac{3}{a-b}=\\=\cfrac{3a(a+b)}{(a-b)(a+b)}-\cfrac{3a}{(a-b)(a+b)}\cdot \cfrac{ab}{a+b}+\cfrac{3}{a-b}=\cfrac{3a+3}{a-b}-\cfrac{3a^2b}{(a-b)(a+b)^2}[/tex]
Второе преобразование
[tex]\left ( \cfrac{3a+3}{a-b}-\cfrac{3a^2b}{(a-b)(a+b)^2} \right ):\cfrac{2a+b}{a^2+2ab+b^2}=\left ( \cfrac{3a+3}{a-b}-\cfrac{3a^2b}{(a-b)(a+b)^2} \right )\cdot \cfrac{(a+b)^2}{2a+b}=\\=\cfrac{3a+3}{a-b}\cdot \cfrac{(a+b)^2}{2a+b}-\cfrac{3a^2b}{(a-b)(a+b)^2}\cdot \cfrac{(a+b)^2}{2a+b}=\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}-\cfrac{3a^2b}{(a-b)(2a+b)}[/tex]
Третье преобразование
[tex]\left (\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}-\cfrac{3a^2b}{(a-b)(2a+b)} \right ):\cfrac{3}{a+b}=\\=\left (\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}-\cfrac{3a^2b}{(a-b)(2a+b)} \right )\cdot \cfrac{a+b}{3}=\\=\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}\cdot \cfrac{a+b}{3}-\cfrac{3a^2b}{(a-b)(2a+b)}\cdot \cfrac{a+b}{3}=\cfrac{(a+1)(a+b)^3}{(a-b)(2a+b)}-\cfrac{a^2b(a+b)}{(a-b)(2a+b)}[/tex]
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Answers & Comments
[tex]\left (\left (\cfrac{3a}{a^2-b^2}\cdot \cfrac{a^2+b^2+ab}{a+b}-\cfrac{3}{b-a} \right ):\cfrac{2a+b}{a^2+2ab+b^2} \right ):\cfrac{3}{a+b}[/tex]
Первое преобразование
[tex]\cfrac{3a}{a^2-b^2}\cdot \cfrac{a^2+b^2+ab}{a+b}-\cfrac{3}{b-a}=\cfrac{3a}{(a-b)(a+b)}\cdot \cfrac{a^2+2ab+b^2-ab}{a+b}+\cfrac{3}{a-b}=\\=\cfrac{3a}{(a-b)(a+b)}\cdot \cfrac{(a+b)^2-ab}{a+b}+\cfrac{3}{a-b}=\\=\cfrac{3a}{(a-b)(a+b)}\left ( a+b-\cfrac{ab}{a+b} \right )+\cfrac{3}{a-b}=\\=\cfrac{3a(a+b)}{(a-b)(a+b)}-\cfrac{3a}{(a-b)(a+b)}\cdot \cfrac{ab}{a+b}+\cfrac{3}{a-b}=\cfrac{3a+3}{a-b}-\cfrac{3a^2b}{(a-b)(a+b)^2}[/tex]
Второе преобразование
[tex]\left ( \cfrac{3a+3}{a-b}-\cfrac{3a^2b}{(a-b)(a+b)^2} \right ):\cfrac{2a+b}{a^2+2ab+b^2}=\left ( \cfrac{3a+3}{a-b}-\cfrac{3a^2b}{(a-b)(a+b)^2} \right )\cdot \cfrac{(a+b)^2}{2a+b}=\\=\cfrac{3a+3}{a-b}\cdot \cfrac{(a+b)^2}{2a+b}-\cfrac{3a^2b}{(a-b)(a+b)^2}\cdot \cfrac{(a+b)^2}{2a+b}=\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}-\cfrac{3a^2b}{(a-b)(2a+b)}[/tex]
Третье преобразование
[tex]\left (\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}-\cfrac{3a^2b}{(a-b)(2a+b)} \right ):\cfrac{3}{a+b}=\\=\left (\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}-\cfrac{3a^2b}{(a-b)(2a+b)} \right )\cdot \cfrac{a+b}{3}=\\=\cfrac{(3a+3)(a+b)^2}{(a-b)(2a+b)}\cdot \cfrac{a+b}{3}-\cfrac{3a^2b}{(a-b)(2a+b)}\cdot \cfrac{a+b}{3}=\cfrac{(a+1)(a+b)^3}{(a-b)(2a+b)}-\cfrac{a^2b(a+b)}{(a-b)(2a+b)}[/tex]