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A = \(\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}\)
<=> A3 = 20 - 3×2A
<=> A3 + 6A - 20 = 0
<=> A = 2
\(F=\sqrt[3]{27-27\sqrt{2}+18-2\sqrt{2}}\)\(+\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}\)
\(F=\sqrt[3]{\left(3-\sqrt{2}\right)^3}+\sqrt[3]{\left(3+\sqrt{2}\right)^3}\)
\(F=3+\sqrt{2}+3-\sqrt{2}=6\)
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)
C= 3√45+29√2+3√45−29√2
⇔\(C^3=45+29\sqrt{2}+45-29\sqrt{2}+3\sqrt[3]{45+29\sqrt{2}}.\sqrt[3]{45-29\sqrt{2}}\left(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\right)\\ C^3=90+3\sqrt[3]{343}.C\\ C^3=90+21C\\ C^3-21C-90=0\\ C^3-36C+15C-90\\ C\left(C-6\right)\left(C+6\right)+15\left(C-6\right)=0\\ \left(C-6\right)\left[C\left(C+6\right)+15\right]=0\\ \left(C-6\right)\left(C^2+6C+15\right)=0\\ \)
Mà C2+6C+15=(C+3)2+6 > 0
Nên C-6=0
⇒C=6
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: \(=\dfrac{\sqrt{23-8\sqrt{7}}}{3}+\dfrac{\sqrt{23+8\sqrt{7}}}{3}\)
\(=\dfrac{4-\sqrt{7}+4+\sqrt{7}}{3}=\dfrac{8}{3}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3}+9}=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}=\)
\(\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1=1.\)
\(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
\(=\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}+\sqrt[3]{27-27\sqrt{2}+18-2\sqrt{2}}\)
\(=\sqrt[3]{\left(3+\sqrt{2}\right)^3}+\sqrt[3]{\left(3-\sqrt{2}\right)^3}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)