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: căn 6+2căn 5-căn 5

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

b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)

=>a^3-3a-4=0

=>a^3-3a=4

\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)

\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)

=4

a: Sửa đề: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\cdot\left(\sqrt{3}-1\right)}\)

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

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

b: \(\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt[3]{\left(\sqrt{3}\right)^3+3\cdot\left(\sqrt{3}\right)^2\cdot1+3\cdot\sqrt{3}\cdot1^2+1^3}}\)

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

\(=\sqrt{3+\sqrt{3}+\sqrt{3}+1}\)

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

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

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

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

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

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

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

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

\(C=3\)

sao gặp toàn bài lp 9 z ????

\(=\frac{2\sqrt{5\cdot3}-2\sqrt{5\cdot2}+\sqrt{2\cdot3}-\sqrt{3\cdot3}}{2\sqrt{5}-2\sqrt{2\cdot5}-\sqrt{3}+\sqrt{2\cdot3}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(2\sqrt{5}-\sqrt{3}\right)}{-\left(\sqrt{2}-1\right)\cdot\left(2\sqrt{5}-\sqrt{3}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)

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

\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)

b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)

\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)

c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)

\(=\dfrac{-2\left(3-2\sqrt{2}\right)+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{-6+4\sqrt{2}+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}\)

\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)

\(a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)

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

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

\(=\sqrt{4+2\sqrt{3}}\)

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

\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)

\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)

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

\(=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}=\sqrt{4-2\sqrt{3}}\)

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

\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)

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

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

\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)

\(=\sqrt{\sqrt{3}+5-\sqrt{3}}=\sqrt{5}\)

\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)

\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)

\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)

\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)

\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)

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: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)

\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)

\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)

\(=\sqrt{11+6\sqrt{2}}\)

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