Question

Даю 25 баллов, 6.16 задание. Срочно!​

Answer 1

Ответ:

$1)\ \displaystyle \frac{a-b}{a^{0,5}-b^{0,5}}-\frac{a^{1,5}-b^{1,5}}{a-b}=\frac{(a^{0,5}-b^{0,5})(a^{0,5}+a^{0,5})}{a^{0,5}-b^{0,5}}-\frac{(a^{0,5}-b^{0,5})(a+a^{0,5}b^{0,5}+b)}{(a^{0,5}-b^{0,5})(a^{0,5}+b^{0,5})}=\\\\\\=(a^{0,5}+b^{0,5})-\frac{a+a^{0,5}\, b^{0,5}+b}{a^{0,5}+b^{0,5}}=\frac{a+2a^{0,5}\, b^{0,5}+b-a-a^{0,5}\, b^{0,5}-b}{a^{0,5}+b^{0,5}}=\frac{a^{0,5}\, b^{0,5}}{a^{0,5}+b^{0,5}}$

$\displaystyle 2)\ \frac{m-n}{m^{\frac{1}{3}}-n^{\frac{1}{3}}}-\frac{m+n}{m^{\frac{1}{3}}+n^{\frac{1}{3}}}=\\\\\\=\frac{(m^{\frac{1}{3}}-n^{\frac{1}{3}})(m^{\frac{2}{3}}+m^{\frac{1}{3}}\, n^{\frac{1}{3}}+n^{\frac{2}{3}})}{m^{\frac{1}{3}}-n^{\frac{1}{3}}}-\frac{(m^{\frac{1}{3}}+n^{\frac{1}{3}})(m^{\frac{2}{3}}-m^{\frac{1}{3}}\, n^{\frac{1}{3}}+n^{\frac{2}{3}})}{m^{\frac{1}{3}}+n^{\frac{1}{3}}}=$

$=m^{\frac{2}{3}}+m^{\frac{1}{3}}\, n^{\frac{1}{3}}+n^{\frac{2}{3}}-(m^{\frac{2}{3}}+m^{\frac{1}{3}}\, n^{\frac{1}{3}}+n^{\frac{2}{3}})=2m^{\frac{1}{3}}\, n^{\frac{1}{3}}$