Question

решите прошу! отдаю последние баллы

Answer 1

Решение на фотографии.

Answer 2

$1) \ \Big(1-\dfrac{a-b}{a+b} \Big)\Big(2+\dfrac{2b}{a-b}\Big)=\dfrac{a+b-a+b}{a+b} \cdot\dfrac{2a-2b+2b}{a-b} =\\\\\\=\dfrac{2b}{a+b} \cdot\dfrac{2a}{a-b} =\boxed{\dfrac{4ab}{a^{2}-b^{2}} }$

$2) \ \Big(1+\dfrac{a+b}{a-b} \Big)\Big(2-\dfrac{2a}{a+b} \Big)=\dfrac{a-b+a+b}{a-b} \cdot\dfrac{2a+2b-2a}{a+b}=\\\\\\=\dfrac{2a}{a-b}\cdot\dfrac{2b}{a+b} =\boxed{\dfrac{4ab}{a^{2} -b^{2} } }$

$3) \ \Big(\dfrac{6}{a-b} -\dfrac{5}{a+b} \Big)\cdot\dfrac{a-b}{a+11b}=\dfrac{6a+6b-5a+5b}{(a-b)(a+b)} \cdot\dfrac{a-b}{a+11b}=\\\\\\=\dfrac{a+11b}{a+b} \cdot\dfrac{1}{a+11b} =\boxed{\dfrac{1}{a+b}}$

$4) \ \Big(\dfrac{3}{c}+\dfrac{3}{c+d} \Big)\cdot\dfrac{c}{18(2c+d)} =\dfrac{3c+3d+3c}{c(c+d)} \cdot\dfrac{c}{18(2c+d)} =\\\\=\dfrac{6c+3d}{c(c+d)} \cdot\dfrac{c}{18(2c+d)} =\dfrac{3(2c+d)}{c(c+d)} \cdot\dfrac{c}{18(2c+d)} =\boxed{\dfrac{1}{6(c+d)} }$