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a) Đặt \(A=3+\sqrt{3}\)
<=>\(A^3=27+27\sqrt{3}+27+3\sqrt{3}\)
<=>\(A^3=54+30\sqrt{3}\)
<=>\(A=\sqrt[3]{54+30\sqrt{3}}\)
Vậy....
b) mình sửa lại đề nhá:
Tính \(B=\sqrt[3]{54+30\sqrt{3}}+\sqrt[3]{54-30\sqrt{3}}\)
\(B=\sqrt[3]{\left(3+\sqrt{3}\right)^3}+\sqrt[3]{\left(3-\sqrt{3}\right)^3}\)
\(B=3+\sqrt{3}+3-\sqrt{3}=6\)
Ta có : \(\sqrt{54-14\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{7^2-2.7.\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{3^2+2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(7-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(=7-\sqrt{5}-\left(3+\sqrt{5}\right)=7-\sqrt{5}-3-\sqrt{5}=4\)
a) \(\frac{x\sqrt[3]{y}+\sqrt[3]{x^2y^2}}{\sqrt[3]{x^2y^2}+y\sqrt[3]{x}}\)
\(=\frac{\sqrt[3]{x^2y}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)}{\sqrt[3]{xy^2}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)}=\sqrt[3]{\frac{x^2y}{xy^2}}=\sqrt[3]{\frac{x}{y}}\)
b) \(\frac{\sqrt[3]{54}-2\sqrt[3]{16}}{\sqrt[3]{54}+2\sqrt[3]{16}}\)
\(=\frac{\sqrt[3]{27.2}-2\sqrt[3]{8.2}}{\sqrt[3]{27.2}+2\sqrt[3]{8.2}}\)
\(=\frac{3\sqrt[3]{2}-4\sqrt[3]{2}}{3\sqrt[3]{2}+4\sqrt[3]{2}}=\frac{-\sqrt[3]{2}}{7\sqrt[3]{2}}=-\frac{1}{7}\)
\(A=\sqrt[3]{\dfrac{384}{3}}+3\cdot\left(-3\right)\cdot\sqrt[3]{2}+6\sqrt[3]{2}\)
\(=4\sqrt[3]{2}-9\sqrt[3]{2}+6\sqrt[3]{2}\)
\(=\sqrt[3]{2}\)
\(5\sqrt[3]{2}+\sqrt[3]{-16}+\sqrt[3]{54}=5\sqrt[3]{2}-2\sqrt[3]{2}+3\sqrt[3]{2}=6\sqrt[3]{2}\)
\(5\sqrt[3]{2}+\sqrt[3]{-16}+\sqrt[3]{54}\)
\(=5\sqrt[3]{2}-2\sqrt[3]{2}+3\sqrt[3]{2}\)
\(=6\sqrt[3]{2}\)
\(=\sqrt{2008}\)
\(=2\sqrt{502}\)
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trả lời
\(\sqrt{1574+54}\)
\(=\sqrt{2008}\)
\(=2\sqrt{502}\)