Question

СРОЧНО!!! ДАЮ 25 БАЛЛОВ!!!​

Answer 1

Ответ:   1 .

$\dfrac{1}{\sqrt2+1}+\dfrac{1}{\sqrt3+\sqrt2}+\dfrac{1}{\sqrt4+\sqrt3}=\\\\\\=\dfrac{\sqrt2-1}{(\sqrt2+1)(\sqrt2-1)}+\dfrac{\sqrt3-\sqrt2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}+\dfrac{\sqrt4-\sqrt3}{(\sqrt4+\sqrt3)(\sqrt4-\sqrt3)}=\\\\\\=\dfrac{\sqrt2-1}{2-1}+\dfrac{\sqrt3-\sqrt2}{3-2}+\dfrac{\sqrt4-\sqrt3}{4-3}=\\\\\\=(\sqrt2-1)+(\sqrt3-\sqrt2)+(\sqrt4-\sqrt3)=-1+\sqrt4=-1+2=1$

Answer 2

$\frac{1}{ \sqrt{2} + 1 } + \frac{1}{ \sqrt{3} + \sqrt{2} } + \frac{1}{ \sqrt{4} + \sqrt{3} } = \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \frac{1}{ 2 + \sqrt{3} } = \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} = 1$