3,4,5,6 примеры все варианты

Вариант №2

Задача 3.
$b^{\frac{3}{4}}+b^{\frac{1}{2}}=b^{\frac{3}{4}}+b^{\frac{2}{4}}=$

$=b^{\frac{2}{4}}\left(b^{\frac{1}{4}}+1\right)=b^{\frac{1}{2}}\left(b^{\frac{1}{4}}+1\right)$

Задача 4.
$\frac{x^{\frac{3}{2}}-y^{\frac{3}{2}}}{x+x^{\frac{1}{2}}y^{\frac{1}{2}}+y}=$

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

$=\frac{\left(x^{\frac{1}{2}}-y^{\frac{1}{2}}\right)\left(\left(x^{\frac{1}{2}}\right)^2+x^{\frac{1}{2}}y^{\frac{1}{2}}+\left(y^{\frac{1}{2}}\right)^2\right)}{\left(x^{\frac{1}{2}}\right)^2+x^{\frac{1}{2}}y^{\frac{1}{2}}+\left(y^{\frac{1}{2}}\right)^2}=$

$=x^{\frac{1}{2}}-y^{\frac{1}{2}}$

Задача 5.
$\sqrt[5]{\left(1\frac{1}{4}-1\frac{1}{5}\right)^{3}}=\left(\frac{5}{4}-\frac{6}{5}\right)^{\frac{3}{5}}=$

$=\left(\frac{25-24}{20}\right)^{\frac{3}{5}}=\left(\frac{1}{20}\right)^{\frac{3}{5}}=$

$\sqrt[5]{\left(1\frac{1}{6}-1\frac{1}{7}\right)^{3}}=\left(\frac{7}{6}-\frac{8}{7}\right)^{\frac{3}{5}}=$

$=\left(\frac{49-48}{42}\right)^{\frac{3}{5}}=\left(\frac{1}{42}\right)^{\frac{3}{5}}=$

$\left(\frac{1}{20}\right)^{\frac{3}{5}}>\left(\frac{1}{42}\right)^{\frac{3}{5}}$

Задача 6.

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

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

Вариант №1 - на фото.