Question

решите пожалуйста(◍•ᴗ•◍)❤ ​

Answer 1

Объяснение:

$\displaystyle \left(\frac{\sqrt[3]{b} }{\sqrt[6]{b}+\sqrt[6]{a} } -\frac{\sqrt{b} }{\sqrt[3]{b}+2\sqrt[6]{ab}+\sqrt[3]{a} } \right):\left(\frac{\sqrt[6]{b} }{\sqrt[6]{b} -\sqrt[6]{a} } -\frac{\sqrt[3]{b} }{\sqrt[3]{b}-\sqrt[3]{a} } \right)=\\\\= \left(\frac{\sqrt[3]{b} }{\sqrt[6]{b}+\sqrt[6]{a} } -\frac{\sqrt{b} }{(\sqrt[6]{b}+\sqrt[6]{a} )^2 } \right):\left(\frac{\sqrt[6]{b} }{\sqrt[6]{b} -\sqrt[6]{a} } -\frac{\sqrt[3]{b} }{(\sqrt[6]{b}-\sqrt[6]{a})(\sqrt[6]{b}+\sqrt[6]{a}) } \right)=$

$\displaystyle =\frac{\sqrt[3]{b} *(\sqrt[6]{b}+\sqrt[6]{a})-\sqrt{b} }{(\sqrt[6]{b}+\sqrt[6]{a})^2 } :\frac{\sqrt[6]{b}*(\sqrt[6]{b}+\sqrt[6]{a})-\sqrt[3]{b} }{(\sqrt[6]{b}-\sqrt[6]{a})(\sqrt[6]{b}+\sqrt[6]{a}) } =$

$\displaystyle = \frac{\sqrt[6]{b^2b}+\sqrt[6]{ab^2}-\sqrt{b} }{(\sqrt[6]{b}+\sqrt[6]{a})^2 } *\frac{(\sqrt[6]{b}-\sqrt[6]{a})(\sqrt[6]{b}+\sqrt[6]{a}) }{\sqrt[6]{b^2}+\sqrt[6]{ab}-\sqrt[3]{b} } =$

$\displaystyle =\frac{\sqrt{b}+\sqrt[6]{ab^2}-\sqrt{b} }{(\sqrt[6]{b}+\sqrt[6]{a})^2 } *\frac{(\sqrt[6]{b}-\sqrt[6]{a})(\sqrt[6]{b}+\sqrt[6]{a})}{\sqrt[3]{b}+\sqrt[6]{ab}-\sqrt[3]{b} } =$

$\displaystyle =\frac{\sqrt[6]{ab^2} *(\sqrt[6]{b}-\sqrt[6]{a}) }{(\sqrt[6]{b}+\sqrt[6]{a}) *\sqrt[6]{ab} } =\frac{\sqrt[6]{b}(\sqrt[6]{b} -\sqrt[6]{a}) }{ \sqrt[6]{b} +\sqrt[6]{a} }$

Использовали формулы:

a²-b²=(a-b)(a+b)

(a+b)² = a²+2ab+b²