Сократить дробь1)(100+2xy-x^2-y^2)/10x+10y+x^2-y^2) 2)(a^2-ab-bc-c^2)/b^2-a^2+2ac-c^2)...

Решите задачу:

$\frac{100+2xy-x^2-y^2}{10x+10y+x^2-y^2} = \frac{100-(x^2-2xy+y^2)}{(10x+10y)+(x^2-y^2)} = \frac{10^2-(x-y)^2}{10(x+y)+(x-y)(x+y)} =$

$=\frac{(10-(x-y))(10+(x-y))}{(x+y)(10+x-y)} =\frac{(10-x+y)(10+x-y)}{(x+y)(10+x-y)} =\frac{10-x+y}{x+y}$

$\frac{a^2-ab-bc-c^2}{b^2-a^2+2ac-c^ 2}= \frac{(a^2-c^2)+(-ab-bc)}{b^2-(a^2-2ac+c^ 2)}= \frac{(a-c)(a+c)-b(a+c)}{b^2-(a-c)^ 2}=$

$= \frac{(a+c)(a-c-b)}{(b-(a-c))(b+(a-c))}= \frac{(a+c)(a-c-b)}{(b-a+c)(b+a-c)}=\frac{(a+c)(a-c-b)}{-(a-b-c)(b+a-c)}=\frac{a+c}{-(b+a-c)}=$

$=\frac{a+c}{c-b-a}$

$\frac{25-a^2 -2ab-b^2}{a^2+ab+5b-25} = \frac{5^2-(a^2+2ab+b^2)}{(a^2-25)+(ab+5b)} =\frac{5^2-(a+b)^2}{(a-5)(a+5)+b(a+5)} =$

$=\frac{(5-(a+b))(5+a+b)}{(a+5)(a-5+b)} =\frac{(5-a-b)(5+a+b)}{-(a+5)(5-a-b)} =\frac{5+a+b}{-(a+5)} =\frac{5+a+b}{-a-5}$