Question

Срочно! Упростить выражение ​

Answer 1

$(\sqrt[4]{m}+\sqrt[4]{n})^2=\sqrt{m}+\sqrt[4]{mn}+\sqrt{n}\\ \\(\sqrt[4]{m}-\sqrt[4]{n})^2=\sqrt{m}-\sqrt[4]{mn}+\sqrt{n}\\ \\ (\sqrt[4]{m}+\sqrt[4]{n})^2+(\sqrt[4]{m}-\sqrt[4]{n})^2=\sqrt{m}+\sqrt[4]{mn}+\sqrt{n}+\sqrt{m}-\sqrt[4]{mn}+\sqrt{n}=2\sqrt{m}+2\sqrt{n}$

$m-n=(\sqrt{m}-\sqrt{n})(\sqrt{m}+\sqrt{n} )$

$\sqrt{m^3}-\sqrt{n^3}=(\sqrt{m})^3-(\sqrt{n})^3 =(\sqrt{m}-\sqrt{n})((\sqrt{m})^2+\sqrt{m}\sqrt{n}+(\sqrt{n})^2)=(\sqrt{m}-\sqrt{n})(m-\sqrt{mn}+n)$

Итак

$\frac{(\sqrt[4]{m}+\sqrt[4]{n})^2+(\sqrt[4]{m}-\sqrt[4]{n})^2}{2(m-n)}:\frac{1}{\sqrt{m^3}-\sqrt{n^3}}-3\sqrt{mn}=\\ \\=\frac{2(\sqrt{m}+\sqrt{n)}}{2(\sqrt{m}-\sqrt{n)}(\sqrt{m}+\sqrt{n})}\cdot\frac{(\sqrt{m}-\sqrt{n})(m+\sqrt{mn}+n)}{1}-3\sqrt{mn}= m-2\sqrt{mn}+n=\\\\=(\sqrt{m}-\sqrt{n})^2$