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a: \(2\sqrt{45}+\sqrt{5}-3\sqrt{80}\)
\(=6\sqrt{5}+\sqrt{5}-12\sqrt{5}\)
\(=-5\sqrt{5}\)
b: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}\)
\(=-8\sqrt{3}+1\)
a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)
\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)
\(=\dfrac{5}{2}\sqrt{5}+7\)
b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)
\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)
\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)
\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)
\(=\dfrac{1}{\sin25^0}-1\)
\(=\dfrac{1-\sin25^0}{\sin25^0}\)
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
2: \(=\sqrt{2}-1-\sqrt{2}=-1\)
3: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
4: \(=1+\dfrac{2-\sqrt{3}}{2-\sqrt{3}}=1+1=2\)
a: \(=\sqrt{7+\sqrt{45}}-\sqrt{7-\sqrt{45}}\)
\(=\dfrac{\sqrt{14+2\sqrt{45}}-\sqrt{14-2\sqrt{45}}}{\sqrt{2}}\)
\(=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
b: \(=2\cdot\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(2\cdot4\sqrt{5}-2\sqrt{5}-3\cdot2\sqrt{5}\right)\)
\(=\sqrt{\sqrt{3}}\cdot0=0\)
c: \(=\left(2-\sqrt{3}-6+2\sqrt{3}+8+2\sqrt{3}\right)\left(4-3\sqrt{3}\right)\)
\(=\left(4+3\sqrt{3}\right)\left(4-3\sqrt{3}\right)\)
=16-27=-11
a ) \(2\sqrt{45}+\sqrt{5}-3\sqrt{80}\)
= \(2\sqrt{9.5}+\sqrt{5}-3\sqrt{16.5}\) \
= \(2.3\sqrt{5}+\sqrt{5}-3.4\sqrt{5}\)
= \(6\sqrt{5}+\sqrt{5}-12\sqrt{5}\)
= \(\left(6+1-12\right)\sqrt{5}\)
= \(-5\sqrt{5}\)
b ) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}+1}-6\sqrt{\dfrac{16}{3}}\)
= / \(2-\sqrt{3}\) / \(+\dfrac{2.\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}-6\sqrt{\dfrac{48}{3^2}}\)
= \(2-\sqrt{3}+\dfrac{2.\left(\sqrt{3}-1\right)}{\sqrt{3}^2-1^2}-\dfrac{6}{3}\sqrt{48}\)
= \(2-\sqrt{3}+\dfrac{2.\left(\sqrt{3}-1\right)}{3-1}-2\sqrt{48}\)
=\(2-\sqrt{3}+\sqrt{3}-1-2\sqrt{16.3}\)
= \(2-\sqrt{3}+\sqrt{3}-1-8\sqrt{3}\)
= \(1-8\sqrt{3}\)
ý c ) em không biết làm ☹