Question

Помогите плиззззззззззззззззз очень надо

Answer 1

$\displaystyle\tt{} \ \ \ \bigg(\frac{2ck}{c^2-k^2}+\frac{c-k}{2c+2k}\bigg)\cdot\frac{20c}{c+k}+\frac{10k}{k-c}=\\\\\\ = \bigg(\frac{2ck}{(c-k)(c+k)}+\frac{c-k}{2(c+k)}\bigg)\cdot\frac{20c}{c+k}-\frac{10k}{c-k}=\\\\\\ = \frac{2ck\cdot2+(c-k)^2}{2(c-k)(c+k)} \cdot\frac{20c}{c+k}-\frac{10k}{c-k}=\\\\\\=\frac{4ck+c^2-2ck+k^2}{(c-k)(c+k)} \cdot\frac{10c}{c+k}-\frac{10k}{c-k}=\\\\\\=\frac{c^2+2ck+k^2}{(c-k)(c+k)} \cdot\frac{10c}{c+k}-\frac{10k}{c-k}=$

$\displaystyle\tt=\frac{(c+k)^2}{(c-k)(c+k)} \cdot\frac{10c}{c+k}-\frac{10k}{c-k}=\frac{10c}{c-k}-\frac{10k}{c-k}=\frac{10(c-k)}{c-k}= 10$

Ответ: 10.