Question

Помогите пожалуйста ​

Answer 1

$\displaystyle \frac{10d}{d^2-n^2} +\frac{10}{n-d} =? \ \ ; \ \ pri \ \ d=-3 \ \ , \ \ n=-1 \\\\\\\frac{10d}{(d-n)(d+n)} +\frac{10}{(n-d)} =10 \bigg(\frac{d}{(d-n)(d+n)} - \frac{1}{d-n} ^{/d+n}\bigg) = \\\\\\ 10\bigg (\frac{d\!\!\! \backslash-d\!\!\!\backslash-n }{(d-n)(d+n)} \bigg)=\frac{-10n}{d^2-n^2} =\frac{-1\cdot (-10)}{(-3)^2-(-1)^2} =\frac{10}{8} =\boxed{\frac{5}{4} }$