Question

Помогите плиззззззззззззззззз очень надо

Answer 1

$displaystylett{} bigg(frac{2ck}{c^2-k^2}+frac{c-k}{2c+2k}bigg)cdotfrac{20c}{c+k}+frac{10k}{k-c}=\\\ = bigg(frac{2ck}{(c-k)(c+k)}+frac{c-k}{2(c+k)}bigg)cdotfrac{20c}{c+k}-frac{10k}{c-k}=\\\ = frac{2ckcdot2+(c-k)^2}{2(c-k)(c+k)} cdotfrac{20c}{c+k}-frac{10k}{c-k}=\\\=frac{4ck+c^2-2ck+k^2}{(c-k)(c+k)} cdotfrac{10c}{c+k}-frac{10k}{c-k}=\\\=frac{c^2+2ck+k^2}{(c-k)(c+k)} cdotfrac{10c}{c+k}-frac{10k}{c-k}=$

$displaystylett=frac{(c+k)^2}{(c-k)(c+k)} cdotfrac{10c}{c+k}-frac{10k}{c-k}=frac{10c}{c-k}-frac{10k}{c-k}=frac{10(c-k)}{c-k}= 10$

Ответ: 10.