\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(C=\sqrt{4+5}\)

\(C=3\)

a.

\(\sqrt[3]{125}.\sqrt[3]{\frac{16}{10}}.\sqrt[3]{-0,5}=\sqrt[3]{125.\frac{16}{10}.(-0,5)}=\sqrt[3]{-100}\)

b.

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

c.

\(\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}=\sqrt{3}+\sqrt[3]{(\sqrt{3}+1)^3}=\sqrt{3}+\sqrt{3}+1=2\sqrt{3}+1\)

d.

\(\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt[3]{(\sqrt{3}+1)^3}}=\frac{(\sqrt{3}+1)^2}{\sqrt{3}+1}=\sqrt{3}+1\)

e.

Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)

Khi đó:

$a^3+b^3=4$

$ab=\frac{2}{3}$

$E^3=(a+b)^3=a^3+b^3+3ab(a+b)$
$E^3=4+2E$

$E^3-2E-4=0$
$E^2(E-2)+2E(E-2)+2(E-2)=0$

$(E-2)(E^2+2E+2)=0$

Dễ thấy $E^2+2E+2>0$ nên $E-2=0$

$\Leftrightarrow E=2$

a, Sửa đề:

\(A=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-2-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-2+\sqrt{3}}\)

\(=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)

\(=\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}-\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{3}}\)

\(=\dfrac{2\sqrt{2-\sqrt{3}}}{\sqrt{3}}\)

\(=\dfrac{2\sqrt{6-3\sqrt{3}}}{3}\)

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

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

\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)

\(\left(\sqrt{3}+1-\left(\sqrt{3}-1\right)\right)\left(\sqrt{3}+1+\sqrt{3}-1\right)=2\cdot2\sqrt{3}=4\sqrt{3}\)

cách 2 :\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)

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

cách 3 :kết hợp 2 cách trên

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)

\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)

\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)

3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)

=-2

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

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

\(=2-\sqrt{3}+2+\sqrt{3}\)

=4

a) \(\dfrac{1}{3\sqrt{2}-2\sqrt{3}}-\dfrac{1}{2\sqrt{3}+3\sqrt{2}}\)

\(=\dfrac{1}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}\)

\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{\sqrt{6}}\)

\(=\dfrac{2\sqrt{2}}{\sqrt{6}}\)

\(=\dfrac{2\sqrt{3}}{3}\)

b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)

\(=\dfrac{4\left(\sqrt{3}-2\right)}{2\left(\sqrt{3}-2\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

\(=\dfrac{4}{2}-\dfrac{\sqrt{5}+2}{5-4}\)

\(=2-\sqrt{5}-2\)

\(=-\sqrt{5}\)

\(\sqrt{6+2\sqrt{3}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

\(=\sqrt{6+2\sqrt{3}\cdot\sqrt{2-\sqrt{3}}}\)

\(=\sqrt{6+\sqrt{6}\left(\sqrt{3}-1\right)}\)

\(=\sqrt{6+3\sqrt{2}-\sqrt{6}}\)

\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{-1}=-6\\ b,=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)

\(d,=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{18\sqrt{3}}{-19}=\dfrac{-18\sqrt{3}}{19}\\ e,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)