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Bài 1:
Ta có:
\(\left(a-b+c\right)^3=a^3-b^3+c^3-3a^2b+3a^2c+3ab^2+3b^2c+3ac^2-3bc^2-6abc\)
\(\Rightarrow\left(\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\right)^3=\frac{1}{9}-\frac{2}{9}+\frac{4}{9}-\frac{1}{3}.\sqrt[3]{2}+\frac{1}{3}.\sqrt[3]{4}+\frac{1}{3}.\sqrt[3]{4}+\frac{2}{3}.\sqrt[3]{2}\)
\(+\frac{2}{3}.\sqrt[3]{2}-\frac{2}{3}.\sqrt[3]{4}-\frac{4}{3}=\sqrt[3]{2}-1\)
\(\Rightarrow\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}\)
Lời giải:
Đặt \((a^2-1)\sqrt{a^2-4}=m; a^3-3a-2=n\)
Ta thấy:
\(m^2=(a^2-1)^2(a^2-4)=(a-1)^2(a+2)(a+1)^2(a-2)\)
\(=(n+4)n\)
\(M=\frac{n+m}{n+4+m}=\frac{n+\sqrt{n(n+4)}}{n+4+\sqrt{n(n+4)}}\)
\(=\frac{\sqrt{n}(\sqrt{n}+\sqrt{n+4})}{\sqrt{n+4}(\sqrt{n+4}+\sqrt{n})}\)
\(=\sqrt{\frac{n}{n+4}}\)
\(C=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)+\sqrt{a}\left(\sqrt{a}+3\right)-\left(3a+3\right)}{a-9}:\frac{2\sqrt{a}-2-\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\)
\(C=\frac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\frac{\sqrt{a}-3}{2\sqrt{a}-2-\sqrt{a}+3}\)
\(C=\frac{-3\sqrt{a}-3}{\sqrt{a}+3}.\frac{1}{\sqrt{a}+1}\)
\(C=\frac{-3}{\sqrt{a}+3}\)
Thay a = \(21-12\sqrt{3}\) vào C , ta có
\(C=\frac{-3}{\sqrt{21-12\sqrt{3}}+3}\)
\(C=\frac{-3}{\sqrt{\left(2\sqrt{3}-3\right)^2}+3}\)
\(C=\frac{-3}{2\sqrt{3}-3+3}=\frac{-3}{2\sqrt{3}}=\frac{-\sqrt{3}}{2}\)