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1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
\(a.\left(4+\sqrt{7}\right)\left(\sqrt{14}-\sqrt{2}\right)\sqrt{4-\sqrt{7}}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)\sqrt{7-2\sqrt{7}+1}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)^2=2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)=2\left(16-7\right)=18\) \(b.\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}=\dfrac{4\sqrt{2}+\sqrt{14}}{6+\sqrt{7+2\sqrt{7}+1}}+\dfrac{4\sqrt{2}-\sqrt{14}}{6-\sqrt{7-2\sqrt{7}+1}}=\dfrac{4\sqrt{2}+\sqrt{14}}{7+\sqrt{7}}+\dfrac{4\sqrt{2}-\sqrt{14}}{7-\sqrt{7}}=\dfrac{\left(4\sqrt{2}+\sqrt{14}\right)\left(7-\sqrt{7}\right)+\left(4\sqrt{2}-\sqrt{14}\right)\left(7+\sqrt{7}\right)}{49-7}=\dfrac{28\sqrt{2}-4\sqrt{14}+7\sqrt{14}-7\sqrt{2}+28\sqrt{2}+4\sqrt{14}-7\sqrt{14}-7\sqrt{2}}{42}=\dfrac{42\sqrt{2}}{42}=\sqrt{2}\)
a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2\sqrt{3}-1\right|+\left|2-\sqrt{3}\right|\)
\(=2\sqrt{3}-1+2-\sqrt{3}=\sqrt{3}+1\)
b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)
c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)
e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)
f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)
ta có : \(A=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\) \(=\dfrac{4\sqrt{2}}{\sqrt{5}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)
làm tương tự với B rồi --> ...
a) A=12\(\sqrt{3}\)
B= \(\frac{8}{3}\)
c) C= 1
d)...
Chúc bạn học tốt nha ^^!
3: \(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
4: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\sqrt{2}\)
5: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)
\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)