Ответ:
Решение на фотографии.
[tex]\frac{4}{\sqrt{3}-1}= \frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{4(\sqrt{3}+1)}{(\sqrt{3})^2-1^2}=\frac{4(\sqrt{3}+1)}{3-1}=\frac{4(\sqrt{3}+1)}{2}=2(\sqrt{3}+1)=2\sqrt{3}+2[/tex]
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Ответ:
Решение на фотографии.
[tex]\frac{4}{\sqrt{3}-1}= \frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{4(\sqrt{3}+1)}{(\sqrt{3})^2-1^2}=\frac{4(\sqrt{3}+1)}{3-1}=\frac{4(\sqrt{3}+1)}{2}=2(\sqrt{3}+1)=2\sqrt{3}+2[/tex]