Question

2А.21(б) ,2В.35(б),2С.51(б)

Answer 1

$\frac{a^2-1}{4a+8b} \cdot \frac{a^2+4ab+4b^2}{3-3a} = \frac{(a-1)(a+1)}{4(a+2b)} \cdot \frac{(a+2b)^2}{3(1-a)} = \\\ = \frac{(a-1)(a+1)(a+2b)^2}{4(a+2b)\cdot 3(1-a)}=-\frac{(a+1)(a+2b)}{12}$

$(5- \frac{5a+2b}{2a} )\cdot \frac{1}{5a+2b} + \frac{1}{5a-2b}= \frac{10a-5a-2b}{2a}\cdot \frac{1}{5a+2b} + \frac{1}{5a-2b}= \\\ = \frac{5a-2b}{2a}\cdot \frac{1}{5a+2b} + \frac{1}{5a-2b}= \frac{5a-2b}{2a(5a+2b)} + \frac{1}{5a-2b}= \\\ = \frac{(5a-2b)^2+2a(5a+2b)}{2a(5a+2b)(5a-2b)} = \frac{25a^2-20ab+4b^2+10a^2+4ab}{2a(25a^2-4b^2)} = \frac{35a^2-16ab+4b^2}{2a(25a^2-4b^2)}$

$ax^2-ay-bx^2-cy+by+cx^2=x^2(a-b+c)-y(a+c-b)= \\\ =(a-b+c)(x^2-y)$