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\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2}+\sqrt{2+\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2+\sqrt{3}}\right)}\)
\(=\dfrac{2\sqrt{2}}{2-\left(2+\sqrt{3}\right)}=\dfrac{2\sqrt{2}}{-\sqrt{3}}=-\dfrac{2\sqrt{6}}{3}\)
Ta có: \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-2-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-2+\sqrt{3}}\)
\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{-\sqrt{2}-\sqrt{2+\sqrt{3}}}{\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{6}}\)
\(=\dfrac{-2}{\sqrt{6}}=\dfrac{-2\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{3}\)
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
Giải:
Ta có tính chất tổng quát:
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
\(=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng vào biểu thức
\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}\)
\(A=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}}}}}=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}}}\)
\(=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}=\frac{2}{\sqrt{3+1}}=1\)
\(\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{\left(2+\sqrt{2}\right)\left(2^2-2\sqrt{2}+\sqrt{2}^2\right)}{3-\sqrt{2}}-\frac{\sqrt{2}\left(3+\sqrt{2}\right)}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{2\sqrt{2}\left(\sqrt{2}+1\right)\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-3-\sqrt{2}\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=4+2\sqrt{2}-3-\sqrt{2}-2-\sqrt{2}=-1\)
gọi \(A=\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}\)
\(< =>A^2=3+\sqrt{3}+3-\sqrt{3}+2\sqrt{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(< =>A^2=6+2\sqrt{9-3\sqrt{3}+3\sqrt{3}-\sqrt{3^2}}\)
\(< =>A^2=6+2\sqrt{6}\)
\(< =>A=\sqrt{6+2\sqrt{6}}\)