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Ta có: \(\sqrt{3}-3\sqrt{2}=\sqrt{3}-\sqrt{3^2.2}=\sqrt{3}-\sqrt{18}\)
Mà \(3< 18\Leftrightarrow\sqrt{3}< \sqrt{18}\Leftrightarrow\sqrt{3}-\sqrt{18}< 0\)(1)
Lại có: \(5\sqrt{2}-4\sqrt{3}=\sqrt{5^2.2}-\sqrt{4^2.3}=\sqrt{50}-\sqrt{48}\)
Mà \(50>48\Leftrightarrow\sqrt{50}>\sqrt{48}\Leftrightarrow\sqrt{50}-\sqrt{48}>0\)(2)
Từ (1) và (2), ta có: \(\sqrt{3}-3\sqrt{2}< 5\sqrt{2}-4\sqrt{3}\)
\(\frac{5\left(\sqrt{6}-1\right)\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}\)
\(=\frac{5\left(\sqrt{6}-1\right)^2}{5}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{1}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left(\sqrt{6}-1\right)^2-\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{2}-1\right)\)
\(=6-2\sqrt{6}+1-2+2\sqrt{6}-3+\sqrt{2}-1=\sqrt{2}\)
Sủa lại đề:
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có \(a^2+b^2=6\), \(ab=2\), \(a+b=\sqrt{10}\), \(a-b=\sqrt{2}\), \(a^2-b^2=2\sqrt{5}\)
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)
\(=\frac{a^2.\left(\sqrt{10}+b\right)-b^2.\left(\sqrt{10}+a\right)}{\left(\sqrt{10}+a\right).\left(\sqrt{10}+b\right)}\)
\(=\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{\sqrt{10}.\left(a^2-b^2\right)+ab.\left(a-b\right)}{10+\sqrt{10}.\left(a+b\right)+ab}\)
\(=\frac{\sqrt{10}.2\sqrt{5}+\sqrt{10}.\sqrt{2}}{10+\sqrt{10}.\sqrt{10}+2}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}\)
\(=\frac{12\sqrt{2}}{22}\)
\(=\frac{6\sqrt{2}}{11}\)
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}
\)
\(=\frac{3+\sqrt{5}-3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
\(=\frac{0}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
\(=0\)
\(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=|\sqrt{2}-1|=\sqrt{2}-1\)
Tương tự \(\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\); \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
\(\Rightarrow BTT=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\)
\(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{3-2\sqrt{3}+1}-\sqrt{4-4\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{2}-1+\sqrt{3}-1-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}-4\)
Đặt \(a=\sqrt[3]{9+4\sqrt{5}},b=\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow\hept{\begin{cases}a^3+b^3=18\\ab=1\end{cases};a+b=x}\)
Ta có: \(x=a+b\Leftrightarrow x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)\(\Rightarrow x^3=18+3x\Leftrightarrow x^3-3x=18\)(1)
Tương tự: Đặt \(c=\sqrt[3]{3+2\sqrt{2}},d=\sqrt[3]{3-2\sqrt{2}}\)
\(\Rightarrow\hept{\begin{cases}c^3+d^3=6\\cd=1\end{cases};c+d=y}\)
Ta có: \(y=c+d\Leftrightarrow y^3=\left(c+d\right)^3=c^3+d^3+3cd\left(c+d\right)\)\(\Rightarrow y^3=6+3y\)
\(\Leftrightarrow y^3-3y=6\)(2)
Từ (1) và (2) suy ra \(A=x^3-3x+y^3-3y+2020=18+6+2020=2048\)