e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{2}+1-\sqrt{2}+1\)

=2

a)

\(\sqrt[3]{(\sqrt{2}+1)(3+2\sqrt{2})}=\sqrt[3]{(\sqrt{2}+1)(2+2\sqrt{2}+1)}\)

\(=\sqrt[3]{(\sqrt{2}+1)(\sqrt{2}+1)^2}=\sqrt[3]{(\sqrt{2}+1)^3}=\sqrt{2}+1\)

b)

\(\sqrt[3]{(4-2\sqrt{3})(\sqrt{3}-1)}=\sqrt[3]{(3-2\sqrt{3}+1)(\sqrt{3}-1)}\)

\(=\sqrt[3]{(\sqrt{3}-1)^2(\sqrt{3}-1)}=\sqrt[3]{(\sqrt{3}-1)^3}=\sqrt{3}-1\)

c)

\((\sqrt[3]{4}+1)^3-(\sqrt[3]{4}-1)^3=[(\sqrt[3]{4}+1-(\sqrt[3]{4}-1)][(\sqrt[3]{4}+1)^2+(\sqrt[3]{4}+1)(\sqrt[3]{4}-1)+(\sqrt[3]{4}-1)^2]\)

\(=2[\sqrt[3]{16}+1+2\sqrt[3]{4}+\sqrt[3]{16}-1+\sqrt[3]{16}+1-2\sqrt[3]{4}]\)

\(=2(3\sqrt[3]{16}+1)\)

d)

\((\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}+\sqrt[3]{2})=[(\sqrt[3]{3})^2-\sqrt[3]{3}.\sqrt[3]{2}+(\sqrt[3]{2})^2](\sqrt[3]{3}+\sqrt[3]{2})\)

\(=(\sqrt[3]{3})^3+(\sqrt[3]{2})^3=3+2=5\)

e)

\(E=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

Áp dụng công thức $(a+b)^3=a^3+b^3+3ab(a+b)$ ta có:

\(E^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{(20+14\sqrt{2})(20-14\sqrt{2})}.E\)

\(E^3=40+3\sqrt[3]{20^2-(14\sqrt{2})^2}.E\)

\(E^3=40+3\sqrt[3]{8}.E=40+6E\)

\(\Leftrightarrow E^2(E-4)+4E(E-4)+10(E-4)=0\)

\(\Leftrightarrow (E-4)(E^2+4E+10)=0\)

Dễ thấy $E^2+4E+10=(E+2)^2+6\neq 0$ nên $E-4=0$ hay $E=4$

a, \(\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|=2+\sqrt{5}-2+\sqrt{5}=2\sqrt{5}\)

b, \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=\left|\sqrt{3}-1\right|-\left|1+\sqrt{3}\right|=\sqrt{3}-1-1-\sqrt{3}=2\)

c, \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

1: \(\left(\sqrt{15}-2\sqrt3\right)^2+12\sqrt5\)

\(=15+12-2\cdot\sqrt{15}\cdot2\sqrt3+12\sqrt5\)

\(=27-4\sqrt{45}+12\sqrt5=27\)

2: \(3\sqrt2\left(4-\sqrt2\right)+3\left(1-2\sqrt2\right)^2\)

\(=12\sqrt2-6+3\left(9-4\sqrt2\right)\)

\(=12\sqrt2-6+27-12\sqrt2=21\)

3: \(\frac12\left(\sqrt6+\sqrt5\right)^2-\frac14\cdot\sqrt{120}-\sqrt{\frac{15}{2}}\)

\(=\frac12\left(11+2\sqrt{30}\right)-\frac14\cdot2\sqrt{30}-\sqrt{\frac{30}{4}}\)

\(=\frac{11}{2}+\sqrt{30}-\frac12\sqrt{30}-\frac12\sqrt{30}=\frac{11}{2}\)

4: \(\left(\sqrt{4-\sqrt7}-\sqrt{4+\sqrt7}\right)^2\)

\(=4-\sqrt7+4+\sqrt7-2\cdot\sqrt{\left(4-\sqrt7\right)\left(4+\sqrt7\right)}\)

\(=8-2\cdot\sqrt{16-7}=8-2\cdot\sqrt9=8-2\cdot3=8-6=2\)

5: \(\left(\sqrt{\sqrt{14}+\sqrt5}+\sqrt{\sqrt{14}-\sqrt5}\right)^2\)

\(=\sqrt{14}+\sqrt5+\sqrt{14}-\sqrt5+2\cdot\sqrt{\left(\sqrt{14}+\sqrt5\right)\left(\sqrt{14}-\sqrt5\right)}\)

\(=2\sqrt{14}+2\cdot\sqrt{14-5}=2\sqrt{14}+2\cdot\sqrt9=2\sqrt{14}+6\)

6: \(\left(\sqrt3+1\right)^3-\left(\sqrt3-1\right)^3\)

\(=\left(3\sqrt3+3\cdot3\cdot1+3\cdot\sqrt3\cdot1^2+1\right)-\left(3\sqrt3-3\cdot3\cdot1+3\sqrt3-1\right)\)

\(=\left(6\sqrt3+10\right)-\left(6\sqrt3-10\right)=20\)

7: \(\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\)

\(=\left\lbrack\left(\sqrt2\right)^3+3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2+1^3\right\rbrack-\left\lbrack\left(\sqrt2\right)^3-3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2-1^3\right\rbrack\)

\(=\left(2\sqrt2+6+3\sqrt2+1\right)-\left(2\sqrt2-6+3\sqrt2-1\right)=5\sqrt2+7-5\sqrt2+7=14\)

8: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+2\cdot\sqrt{360}}\)

\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt5+5}-\sqrt{45+2\cdot3\sqrt5\cdot2\sqrt2+8}\)

\(=\sqrt{\left(2\sqrt2-\sqrt5\right)^2}-\sqrt{\left(3\sqrt5+2\sqrt2\right)^2}\)

\(=2\sqrt2-\sqrt5-3\sqrt5-2\sqrt2=-4\sqrt5\)

9: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}\right)\)

\(=\frac12\left(\sqrt5-1+\sqrt5+1\right)=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}\)