[TEX]\sqrt[4]{28-16.\sqrt{3}} \\ \sqrt{3} - 1[/TEX]
vì 2 số đều dương ta mũ 4 hai vế
[TEX]VT : 28-16.\sqrt{3} \\ VP = 28 - 16.\sqrt{3} \\ \Rightarrow VP = VT \\ \sqrt[4]{28-16.\sqrt{3}} = \sqrt{3} - 1[/TEX]
câu 2
[TEX]x^3 = 4(\sqrt{5}+1) - 4(\sqrt{5} -1) - 3.\sqrt[3]{16.(1+\sqrt{5})^2}.\sqrt[3]{4(\sqrt{5}-1)} + 3.\sqrt[3]{16.(\sqrt{5}-1)^2}.\sqrt[3]{4(\sqrt{5}+1)} \\ 8 - 3.\sqrt[3]{64.(5-1)(\sqrt{5}+1)} + 3.\sqrt[3]{64.(5-1).(\sqrt{5}-1)} \\ 8 - 12.\sqrt[3]{4.(\sqrt{5}+1)} + 12.\sqrt[3]{4.(\sqrt{5}-1)} = 8 - 12.(\sqrt[3]{4.(\sqrt{5}+1)} -\sqrt[3]{4.(\sqrt{5}-1)})[/TEX]
[TEX]x = 12.(\sqrt[3]{4.(\sqrt{5}+1)} -\sqrt[3]{4.(\sqrt{5}-1)})[/TEX]
vậy
[TEX]x^3 +12x - 8 = 8 - 12.(\sqrt[3]{4.(\sqrt{5}+1)} -\sqrt[3]{4.(\sqrt{5}-1)}) + 12.(\sqrt[3]{4.(\sqrt{5}+1)} -\sqrt[3]{4.(\sqrt{5}-1)}) - 8 = 0 [/TEX]