Даю много БАЛЛОВ решите подробно,а не просто ответ.СРОЧНО.

Решите задачу:

$\displaystyle 1.\\a).\ \ \frac{m+3}{n}+\frac{2-m}{n}=\frac{m+3+2-m}{n}= \frac{5}{n};\\\\b).\ \ \frac{2a+7}{b}-\frac{7}{b}=\frac{2a+7-7}{b}=\frac{2a}{b};\\\\c).\ \ \frac{5a+3b}{9b}-\frac{5a-4b}{9b}=\frac{5a+3b-5a+4b}{9b}=\frac{7b}{9b}= \frac{7}{9};\\\\d).\ \ \frac{1}{2x}+\frac{2}{3y}=\frac{3y+4x}{6xy};\\\\\\ 2. \ \ \frac{(x+3)^{2}}{x^{2}-25}*\frac{x-5}{x^{2}+6x+9}=\frac{(x+3)^{2}(x-5)}{(x-5)(x+5)(x+3)^{2}}=\frac{1}{x+5}\\\\npu\ \ x=15:\\\\ \frac{1}{x+5}= \frac{1}{20}=0,05$

$\displaystyle 3.\ \ \frac{b^{2}}{a^{2}-ab}= \frac{a}{a-b}- \frac{b}{a}-1 \\ \\ \frac{b^{2}}{a(a-b)}- \frac{a^{2}-b(a-b)}{a(a-b)}=-1 \\ \\ \frac{b^{2}-a^{2}+ab-b^{2}}{a(a-b)}=-1 \\ \\ -\frac{a(a-b)}{a(a-b)}=-1 \\ \\ -1=-1 \\ \\ \\ 4.\ \ \frac{4a^{2}}{a-3b}: \frac{2a^{2}}{5a^{2}-15ab}= \frac{4a^{2}}{a-3b}* \frac{5a(a-3b)}{2a^{2}}=10a$