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Chứng minh với mọi số nguyên dương, ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\text{[}\left(n+1\right)\sqrt{n}\text{]}^2-\left(n\sqrt{n+1}\right)^2}\)\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\text{ }\left(n+1\right)^2.n-n^2.\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)n\left(n+1-n\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng: Tính B=....
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\left(\frac{-1}{\sqrt{120}}\right)+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)
Ta có: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow A^3=\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)^3\)
\(=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt[3]{81-80}\cdot A\)
\(=18+3A\)
\(\Rightarrow A^3-3A-18=0\)
\(\Leftrightarrow\left(A^3-3A^2\right)+\left(3A^2-9A\right)+\left(6A-18\right)=0\)
\(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)
Mà \(A^2+3A+6=\left(A+\frac{3}{2}\right)^2+\frac{15}{4}\left(\forall A\right)\)
\(\Rightarrow A=3\)
Vậy A = 3
\(C^3=2-\sqrt{5}+2+\sqrt{5}+3\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=4+3\sqrt[3]{4-5}.C=4-3C\Leftrightarrow C^3+3C-4=0\Leftrightarrow\left(C-1\right)\left(C^2+C+4\right)=0\)
\(\Leftrightarrow C-1=0\Leftrightarrow C=1\)