Question

СРОЧНО!!!!!!!!!!!!!!!!!!ПОМОГИТЕ!!!!!ПОЖАЛУЙСТА!!!!!!!!!!!!!!МНОГО БАЛЛОВ!!!!!!!!!!!!!!!!!ВЫДЕЛЕННОЕ!!!!!!!!!!!!!

Answer 1

в)
$\frac{3 + c}{ {c}^{2} - cd } + \frac{3 + d}{ {d}^{2} - cd } = \frac{3 + c}{c(c - d)} + \frac{3 + d}{d(d - c)} = \frac{3 + c}{c(c - d)} - \frac{3 + d }{d(c - d)} = \frac{d(3 + c) - c(3 + d)}{cd(c - d)} = \frac{3d + cd - 3c - cd}{cd(c - d)} = \frac{3(d - c)}{cd(c - d)} = \frac{ - 3(c - d)}{cd(c - d)} = - \frac{3}{cd}$
г)
$\frac{3m + n}{9 {m}^{2} - 3mn} - \frac{4n}{9 {m}^{2} - {n}^{2} } = \frac{3m + n}{3m(3m - n)} - \frac{4n}{(3m - n)(3m + n)} = \frac{ {(3m + n)}^{2} - 3m \times 4n}{3m(3m - n)(3m + n)} = \frac{9 {m}^{2} + 6mn + {n}^{2} - 12mn }{3m(3m - n)(3m + n)} = \frac{9 {m}^{2} - 6mn + {n}^{2} }{3m(3m - n)(3m + n)} = \frac{ {(3m - n)}^{2} }{3m(3m - n)(3m + n)} = \frac{3m - n}{3 m(3m + n)}$

Answer 2

Решение во вложении. Удачи