Ответ:
Освободиться от иррациональности . Применяем формулу разности квадратов : [tex]\bf (a-b)(a+b)=a^2-b^2[/tex] .
[tex]\bf \displaystyle \frac{1}{2-\sqrt2+\sqrt3-\sqrt6}=\frac{1}{\sqrt2(\sqrt2-1)+\sqrt3(1-\sqrt2)}=\frac{1}{\sqrt2(\sqrt2-1)-\sqrt3(\sqrt2-1)}=\\\\\\=\frac{1}{(\sqrt2-1)(\sqrt2-\sqrt3)}=\frac{(\sqrt2+1)(\sqrt2+\sqrt3)}{(\sqrt2-1)(\sqrt2+1)(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)}=\\\\\\=\frac{(\sqrt2+1)(\sqrt2+\sqrt3)}{(2-1)(2-3)}=\frac{(\sqrt2+1)(\sqrt2+\sqrt3)}{1\cdot (-1)}=-(\sqrt2+1)(\sqrt2+\sqrt3)=\\\\\\=-2-\sqrt6-\sqrt2-\sqrt3[/tex]
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Ответ:
Освободиться от иррациональности . Применяем формулу разности квадратов : [tex]\bf (a-b)(a+b)=a^2-b^2[/tex] .
[tex]\bf \displaystyle \frac{1}{2-\sqrt2+\sqrt3-\sqrt6}=\frac{1}{\sqrt2(\sqrt2-1)+\sqrt3(1-\sqrt2)}=\frac{1}{\sqrt2(\sqrt2-1)-\sqrt3(\sqrt2-1)}=\\\\\\=\frac{1}{(\sqrt2-1)(\sqrt2-\sqrt3)}=\frac{(\sqrt2+1)(\sqrt2+\sqrt3)}{(\sqrt2-1)(\sqrt2+1)(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)}=\\\\\\=\frac{(\sqrt2+1)(\sqrt2+\sqrt3)}{(2-1)(2-3)}=\frac{(\sqrt2+1)(\sqrt2+\sqrt3)}{1\cdot (-1)}=-(\sqrt2+1)(\sqrt2+\sqrt3)=\\\\\\=-2-\sqrt6-\sqrt2-\sqrt3[/tex]