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a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow A^2=2-\sqrt{3}+2+\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=4-2\sqrt{4-3}=4-2=2\)
\(\Rightarrow A=-\sqrt{2}\)
b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)
\(=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow B=\sqrt{2}\)
\(a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)
\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)
\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}\)
\(=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)
\(=\sqrt{\sqrt{3}+5-\sqrt{3}}=\sqrt{5}\)
\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)
\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)
\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)
\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)
\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)
a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
b) Ta có: \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)
\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)
c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{\sqrt{3}+5-\sqrt{3}}\)
\(=\sqrt{5}\)
d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)
\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)
\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)
\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)
\(=\sqrt{11+6\sqrt{2}}\)
\(=3+\sqrt{2}\)
\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{29}\)
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)
c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)
d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)
\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)
e, Ghi đúng đề.
\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)
