Вычислить. 20 баллов тому кто решит
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По действиям.
1) Преобразуем первый множитель:
2) Перемножим первый и третий множители:
3) И, наконец, умножим полученное произведение на второй множитель:
Ответ:
![\sqrt[4]{4-\sqrt{12}}*\sqrt[6]{(1+\sqrt{3})^5}*\sqrt[3]{\sqrt{3}-1}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B4-%5Csqrt%7B12%7D%7D%2A%5Csqrt%5B6%5D%7B%281%2B%5Csqrt%7B3%7D%29%5E5%7D%2A%5Csqrt%5B3%5D%7B%5Csqrt%7B3%7D-1%7D "\sqrt[4]{4-\sqrt{12}}*\sqrt[6]{(1+\sqrt{3})^5}*\sqrt[3]{\sqrt{3}-1}")
![\sqrt[4]{4-\sqrt{12}}=\sqrt[4]{3-\sqrt{4*3}+1}=\sqrt[4]{3-2\sqrt{3}+1}=\\\\=\sqrt[4]{(\sqrt{3})^2-2*\sqrt{3}*1+1^2}=\sqrt[4]{(\sqrt{3}-1)^2}=\sqrt[2]{\sqrt{3}-1}=\\\\=\sqrt{\sqrt{3}-1}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B4-%5Csqrt%7B12%7D%7D%3D%5Csqrt%5B4%5D%7B3-%5Csqrt%7B4%2A3%7D%2B1%7D%3D%5Csqrt%5B4%5D%7B3-2%5Csqrt%7B3%7D%2B1%7D%3D%5C%5C%5C%5C%3D%5Csqrt%5B4%5D%7B%28%5Csqrt%7B3%7D%29%5E2-2%2A%5Csqrt%7B3%7D%2A1%2B1%5E2%7D%3D%5Csqrt%5B4%5D%7B%28%5Csqrt%7B3%7D-1%29%5E2%7D%3D%5Csqrt%5B2%5D%7B%5Csqrt%7B3%7D-1%7D%3D%5C%5C%5C%5C%3D%5Csqrt%7B%5Csqrt%7B3%7D-1%7D "\sqrt[4]{4-\sqrt{12}}=\sqrt[4]{3-\sqrt{4*3}+1}=\sqrt[4]{3-2\sqrt{3}+1}=\\\\=\sqrt[4]{(\sqrt{3})^2-2*\sqrt{3}*1+1^2}=\sqrt[4]{(\sqrt{3}-1)^2}=\sqrt[2]{\sqrt{3}-1}=\\\\=\sqrt{\sqrt{3}-1}")
![\sqrt[]{\sqrt{3}-1}*\sqrt[3]{\sqrt{3}-1}=\\\\=(\sqrt{3}-1)^{1/2}*(\sqrt{3}-1)^{1/3}=\\\\=(\sqrt{3}-1)^{1/2+1/3}=(\sqrt{3}-1)^{5/6}=\sqrt[6]{(\sqrt{3}-1)^5}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Csqrt%7B3%7D-1%7D%2A%5Csqrt%5B3%5D%7B%5Csqrt%7B3%7D-1%7D%3D%5C%5C%5C%5C%3D%28%5Csqrt%7B3%7D-1%29%5E%7B1%2F2%7D%2A%28%5Csqrt%7B3%7D-1%29%5E%7B1%2F3%7D%3D%5C%5C%5C%5C%3D%28%5Csqrt%7B3%7D-1%29%5E%7B1%2F2%2B1%2F3%7D%3D%28%5Csqrt%7B3%7D-1%29%5E%7B5%2F6%7D%3D%5Csqrt%5B6%5D%7B%28%5Csqrt%7B3%7D-1%29%5E5%7D "\sqrt[]{\sqrt{3}-1}*\sqrt[3]{\sqrt{3}-1}=\\\\=(\sqrt{3}-1)^{1/2}*(\sqrt{3}-1)^{1/3}=\\\\=(\sqrt{3}-1)^{1/2+1/3}=(\sqrt{3}-1)^{5/6}=\sqrt[6]{(\sqrt{3}-1)^5}")
![\sqrt[6]{(\sqrt{3}-1)^5}*\sqrt[6]{(\sqrt{3}+1)^5}=\\\\=\sqrt[6]{((\sqrt{3}-1)*{(\sqrt{3}+1))^5}}=\\\\=\sqrt[6]{(\sqrt{3^2}-1^2)^5}=\sqrt[6]{(3-1)^5}=\sqrt[6]{2^{5}}=\sqrt[6]{32}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B%28%5Csqrt%7B3%7D-1%29%5E5%7D%2A%5Csqrt%5B6%5D%7B%28%5Csqrt%7B3%7D%2B1%29%5E5%7D%3D%5C%5C%5C%5C%3D%5Csqrt%5B6%5D%7B%28%28%5Csqrt%7B3%7D-1%29%2A%7B%28%5Csqrt%7B3%7D%2B1%29%29%5E5%7D%7D%3D%5C%5C%5C%5C%3D%5Csqrt%5B6%5D%7B%28%5Csqrt%7B3%5E2%7D-1%5E2%29%5E5%7D%3D%5Csqrt%5B6%5D%7B%283-1%29%5E5%7D%3D%5Csqrt%5B6%5D%7B2%5E%7B5%7D%7D%3D%5Csqrt%5B6%5D%7B32%7D "\sqrt[6]{(\sqrt{3}-1)^5}*\sqrt[6]{(\sqrt{3}+1)^5}=\\\\=\sqrt[6]{((\sqrt{3}-1)*{(\sqrt{3}+1))^5}}=\\\\=\sqrt[6]{(\sqrt{3^2}-1^2)^5}=\sqrt[6]{(3-1)^5}=\sqrt[6]{2^{5}}=\sqrt[6]{32}")
![\sqrt[4]{4-\sqrt{12}}*\sqrt[6]{(1+\sqrt{3})^5}*\sqrt[3]{\sqrt{3}-1}=\sqrt[6]{32}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B4-%5Csqrt%7B12%7D%7D%2A%5Csqrt%5B6%5D%7B%281%2B%5Csqrt%7B3%7D%29%5E5%7D%2A%5Csqrt%5B3%5D%7B%5Csqrt%7B3%7D-1%7D%3D%5Csqrt%5B6%5D%7B32%7D "\sqrt[4]{4-\sqrt{12}}*\sqrt[6]{(1+\sqrt{3})^5}*\sqrt[3]{\sqrt{3}-1}=\sqrt[6]{32}")