Question

86
номер 4 под буквой б
номер 5 под буквой а в​

Answer 1

4б)

$\frac{z-8}{z^{\frac{2}{3}}+2z^{\frac{1}{3}}+4}=\frac{(z^{\frac{1}{3}})^{3}-2^{3}}{z^{\frac{2}{3}}+2z^{\frac{1}{3}}+4}=\frac{(z^{\frac{1}{3}}-2)(z^{\frac{2}{3}}+2z^{\frac{1}{3}}+4)}{z^{\frac{2}{3}}+2z^{\frac{1}{3}}+4}=z^{\frac{1}{3}}-2$

5a)

$\frac{x-y}{x^{\frac{3}{4}}+x^{\frac{1}{2}}y^{\frac{1}{4}}}*\frac{x^{\frac{1}{2}}y^{\frac{1}{4} }+x^{\frac{1}{4}}y^{\frac{1}{2}}}{x^{\frac{1}{2}}+y^{\frac{1}{2}}}=\frac{(x^{\frac{1}{2}}-y^{\frac{1}{2}})(x^{\frac{1}{2}}+y^{\frac{1}{2}})}{x^{\frac{1}{2}}(x^{\frac{1}{4}}+y^{\frac{1}{4}})}*\frac{x^{\frac{1}{4}}y^{\frac{1}{4}}(x^{\frac{1}{4}}+y^{\frac{1}{4}})}{x^{\frac{1}{2}}+y^{\frac{1}{2}}}=\frac{y^{\frac{1}{4}}(x^{\frac{1}{2}}-y^{\frac{1}{2}})}{x^{\frac{1}{4}}}$

5в)

$1)\frac{1}{a+a^{\frac{1}{2}}b^{\frac{1}{2}}}+\frac{1}{a-a^{\frac{1}{2}}b^{\frac{1}{2}}}=\frac{1}{a^{\frac{1}{2}}(a^{\frac{1}{2}}+b^{\frac{1}{2}})}+\frac{1}{a^{\frac{1}{2}}(a^{\frac{1}{2}}-b^{\frac{1}{2}})}=\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}+a^{\frac{1}{2}}+b^{\frac{1}{2}}}{a^{\frac{1}{2}}(a-b)}=\frac{2a^{\frac{1}{2}}}{a^{\frac{1}{2}}(a-b)}=\frac{2}{a-b}$

$2)\frac{2}{a-b}*\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}=\frac{2}{a-b}*\frac{(a-b)(a^{2}+ab+b^{2})}{a^{2}+ab+b^{2}} =2$

Answer 2

ответ: во вложении Объяснение: