Question

алгебра срочно 8 класс​

Answer 1

Ответ:√3-√2

Объяснение:

Answer 2

Ответ:

в) $\sqrt{3}-\sqrt{2}$.

Объяснение:

$\dfrac{\sqrt{2}+1 }{\sqrt{3}+1 } -\dfrac{\sqrt{2}-1 }{\sqrt{3}-1 } ={\dfrac{\sqrt{2}+1}{\sqrt{3}+1 }}^{\setminus (\sqrt{3}-1 )} -{\dfrac{\sqrt{2}-1 }{\sqrt{3}-1 } }^{\setminus (\sqrt{3}+1)} =\\\\\\=\dfrac{(\sqrt{2}+1) \cdot(\sqrt{3}-1 ) }{(\sqrt{3}+1) \cdot(\sqrt{3}-1)}- \dfrac{(\sqrt{2}-1)\cdot(\sqrt{3}+1) }{(\sqrt{3}-1 )\cdot(\sqrt{3}+1)}=\dfrac{(\sqrt{2}+1) \cdot(\sqrt{3}-1 )- (\sqrt{2}-1)\cdot(\sqrt{3}+1) }{(\sqrt{3}+1) \cdot(\sqrt{3}-1)}=$

$=\dfrac{\sqrt{6}-\sqrt{2} +\sqrt{3} -1-\sqrt{6}-\sqrt{2} +\sqrt{3} +1}{3-\sqrt{3} +\sqrt{3} -1} =\dfrac{2\sqrt{3} -2\sqrt{2} }{2} =\dfrac{2(\sqrt{3}-\sqrt{2})}{2}= \\\\\\=\sqrt{3}-\sqrt{2}$