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a ) \(\dfrac{2}{\sqrt{3}-1}\) - \(\dfrac{2}{\sqrt{3}+1}\) = \(\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
= \(\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\) = \(\dfrac{4}{2}\) = 2
b) \(\dfrac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}\) - \(\dfrac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{5\left(2\sqrt{5}-3\sqrt{2}\right)-5\left(2\sqrt{5}+3\sqrt{2}\right)}{12\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12\left(20-18\right)}\)
= \(\dfrac{-30\sqrt{2}}{24}\) = \(\dfrac{-15\sqrt{2}}{12}\) = \(\dfrac{-5\sqrt{2}}{4}\)
c) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\) = \(\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
= \(\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\) = \(\dfrac{60}{20}\) = 3
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)
= \(\dfrac{\sqrt{3}}{\sqrt{2}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)
a/ \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}=2\sqrt{4.2.5\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)
= \(2.2\sqrt{2.5.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}=4.2\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3.2\sqrt{5\sqrt{3}}\)
= \(\sqrt{5\sqrt{3}}\left(8-2-6\right)=\sqrt{5\sqrt{3}}.0=0\)
b/ \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}=2\sqrt{2.4\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)
= \(4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)
c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)
e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
Bài 2:
\(A=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{2}\left(\sqrt{3}+1\right)}=1\)
Bài 1:
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\dfrac{2}{3}\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
a,
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=a-2\sqrt{ab}+b=\left(\sqrt{a}-\sqrt{b}\right)^2\)
b,
A=\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+2\sqrt{12}}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-1-\sqrt{12}}}}{\sqrt{6}+\sqrt{2}}\)\(=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{2}\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)
B=
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)
\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)
\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)
b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
a) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\)
b) \(3\sqrt{12}+5\sqrt{3}-\sqrt{48}=3\sqrt{4.3}+5\sqrt{3}-\sqrt{16.3}=3.2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}+5\sqrt{3}-4\sqrt{3}=7\sqrt{3}\)