Question

решите пожалуйста, очень нужно

Answer 1

Ответ:  $\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\dddddd \displaystyle \Large \boldsymbol{ 0}$

Объяснение:

• $\large \boldsymbol{a-b=(\sqrt{a}-\sqrt{b} )(\sqrt{a}+\sqrt{b} )} \\\\ \boldsymbol$

$b) \ \ \displaystyle \bf \large \boldsymbol{} \frac{a-b}{a-\sqrt{2a} } \cdot \frac{a-2}{\sqrt{a}-\sqrt{b} } \cdot \bigg(\sqrt{2}- \frac{\sqrt{2a} }{\sqrt{a}+\sqrt{2} } \bigg)=\frac{2(\sqrt{a}+\sqrt{b} )}{\sqrt{a} }$

$\displaystyle \large \bf \boldsymbol{} \frac{(\sqrt{a}+\sqrt{b} )(\sqrt{a}-\sqrt{b} )\!\!\!\!\!\!\!\!\!\!\!\!\bigg/}{\sqrt{a}(\sqrt{a} -\sqrt{2})\!\!\!\!\!\!\!\!\!\!\!\!\!\bigg/ } \ \ \cdot \ \ \frac{(\sqrt{a} +\sqrt{2} )(\sqrt{a} -\sqrt{2})\!\!\!\!\!\!\!\!\!\!\!\!\bigg/ }{\sqrt{a}-\sqrt{b} \!\!\!\!\!\!\!\!\!\!\!\!\bigg/ } \ \ \cdot \ \ \bigg(\frac{\sqrt{2a}\!\!\!\!\!\!\diagup -\sqrt{2a}\!\!\!\!\!\!\diagup+2 }{\sqrt{a} +\sqrt{2} } \bigg)=\frac{2(\sqrt{a}+\sqrt{b} )}{\sqrt{a} }$

$\displaystyle \large \bf \boldsymbol{}\boxed{\frac{\sqrt{a}+\sqrt{b} }{\sqrt{a} }} \ \cdot (\sqrt{a} +\sqrt{2} )\!\!\!\!\!\!\!\!\!\!\!\!\bigg/ \ \ \ \cdot \ \frac{2}{(\sqrt{a} +\sqrt{2} )\!\!\!\!\!\!\!\!\!\!\!\!\bigg/} =2\cdot \boxed{\frac{\sqrt{a}+\sqrt{b} }{\sqrt{a} }} \\\\\ 2=2 \\\\2-2=\boxed{0}$