\(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)

\(=\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}+\sqrt[3]{27-27\sqrt{2}+18-2\sqrt{2}}\)

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

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

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(F=\sqrt[3]{27-27\sqrt{2}+18-2\sqrt{2}}\)\(+\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}\)

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

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

C=

⇔\(C^3=45+29\sqrt{2}+45-29\sqrt{2}+3\sqrt[3]{45+29\sqrt{2}}.\sqrt[3]{45-29\sqrt{2}}\left(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\right)\\ C^3=90+3\sqrt[3]{343}.C\\ C^3=90+21C\\ C^3-21C-90=0\\ C^3-36C+15C-90\\ C\left(C-6\right)\left(C+6\right)+15\left(C-6\right)=0\\ \left(C-6\right)\left[C\left(C+6\right)+15\right]=0\\ \left(C-6\right)\left(C^2+6C+15\right)=0\\ \)
Mà C²+6C+15=(C+3)²+6 > 0

Nên C-6=0

⇒C=6

A = \(\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}\)

<=> A³ = 20 - 3×2A

<=> A³ + 6A - 20 = 0

<=> A = 2

2 câu còn lại làm tương tự 

A=2,B=6,C=2

ta có: A³=\(6\sqrt{3}+10-6\sqrt{3}+10-3\sqrt[3]{\left(6\sqrt{3}+10\right)\left(6\sqrt{3}-10\right)}.\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\right)\)

=\(20-3.\sqrt[3]{8}.A\)=\(20-6A\)

do đó A³=20-6A↔A³+6A-20=0↔(A²+2A+10)(A-2)=0

dễ thấy A²+2A+10>0→A=2

b) giống a)

c)giống b)

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: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)

\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)