a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)

\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)

Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)

b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)

\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)

Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)

Lời giải:

Đặt \(\sqrt[3]{27+6\sqrt{21}}=a; \sqrt[3]{27-6\sqrt{21}}=b\) thì ta cần tính tổng $A=a+b$.

Ta có:

$a^3+b^3=54$

\(ab=\sqrt[3]{(27+6\sqrt{21})(27-6\sqrt{21})}=-3\)

$A^3=(a+b)^3=a^3+b^3=3ab(a+b)=54+3(-3)A$

$\Leftrightarrow A^3=54-9A$

$\Leftrightarrow A^3+9A-54=0$

$\Leftrightarrow A^2(A-3)+3A(A-3)+18(A-3)=0$

$\Leftrightarrow (A^2+3A+18)(A-3)=0$

$\Leftrightarrow A-3=0$ (do $A^2+3A+18>0$)

$\Leftrightarrow A=3$

\(E=2\sqrt{3}+3\sqrt{3^3}-\sqrt{100.3}\\ =2\sqrt{3}+9\sqrt{3}-10\sqrt{3}\\ =\left(2+9-10\right)\sqrt{3}=\sqrt{3}\)

\(F=\sqrt{3^2.2}+4\sqrt{18}=\sqrt{18}+4\sqrt{18}=\left(1+4\right)\sqrt{18}=5\sqrt{18}\)

\(G=2\sqrt{3}-4\sqrt{3^3}+5\sqrt{4^2.3}=2\sqrt{3}-12\sqrt{3}+20\sqrt{3}=\left(2-12+20\right)\sqrt{3}=10\sqrt{3}\)

\(H=\left(3\sqrt{25.2}-5\sqrt{9.2}+3\sqrt{2^3}\right)\sqrt{2}\\ =\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\sqrt{2}\\ =6\sqrt{2}.\sqrt{2}=6\)

Ta có  

\(\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)^2\)

\(=27+10\sqrt{2}+27-10\sqrt{2}-2\sqrt{\left(27+10\sqrt{2}\right)\left(27-10\sqrt{2}\right)}\)

\(=54-2\sqrt{529}=8\)

\(\Rightarrow\)  \(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}=\sqrt{8}=2\sqrt{2}\)

Xét tử số

\(\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}\)

\(=\left(\sqrt{27+10\sqrt{2}}.\sqrt{27-10\sqrt{2}}\right)\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)

\(=23\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)

\(=23.2\sqrt{2}=46\sqrt{2}\)

Lại có  \(\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2\)

\(=\sqrt{13}-3+\sqrt{13}+3+2\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)}\)

\(=2\sqrt{13}+2\sqrt{4}=2\sqrt{13}+4\)

ta bình phương mẫu số

\(\left(\frac{\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}}{\sqrt{\sqrt{13}+2}}\right)^2=\frac{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2}{\sqrt{13}+2}\)

\(=\frac{2\sqrt{13}+4}{\sqrt{13}+2}=2\)

Vậy mẫu  \(=\sqrt{2}\)

Vậy  \(x=\frac{46\sqrt{2}}{\sqrt{2}}=46\)  thay vào ta đc A = 92880

Cho biểu thức ban đầu là A
Đặt 3 = a ; \(\sqrt{9+\dfrac{125}{27}}\)= b
⇔A = \(\sqrt[3]{a+b} . \sqrt[3]{b-a}\) 
⇔A=  \(\sqrt[3]{(a+b)(b-a)}\)
⇔A= \(\sqrt[3]{b^2-a^2}\)
⇔A= \(\sqrt[3]{9+\dfrac{125}{27}-9}\)
⇔A= \(\sqrt[3]{\dfrac{125}{27}}\)
⇔A = \(\dfrac{5}{3}\) ( ĐPCM)
 

\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)

=\(\sqrt[3]{-\left(3+\sqrt{9+\dfrac{125}{27}}\right)\left(3-\sqrt{9+\dfrac{125}{27}}\right)}\)

=\(\sqrt[3]{-\left[9-\left(9+\dfrac{125}{27}\right)\right]}\)

=\(\sqrt[3]{\dfrac{125}{27}}\)

=5/3

\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\sqrt{5+3\sqrt{2}}+3\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)}-\dfrac{\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}}{\sqrt{6+3\sqrt{2}}-\sqrt{6-4\sqrt{2}}}\\ =\dfrac{3\left(5+3\sqrt{2}+2\sqrt{25-18}+5-3\sqrt{2}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{\sqrt{4+2+4\sqrt{2}}+\sqrt{4+2-4\sqrt{2}}}{\sqrt{4+2+4\sqrt{2}}-\sqrt{4+2-4\sqrt{2}}}\\ =\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{2+\sqrt{2}+2-\sqrt{2}}{2+\sqrt{2}-2+\sqrt{2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{4}{2\sqrt{2}}=\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)

\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)

\(=\dfrac{\sqrt{9\left(5+3\sqrt{2}\right)}+\sqrt{9\left(5-3\sqrt{2}\right)}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)

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

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

\(=\dfrac{3\left[5+3\sqrt{2}+5-3\sqrt{2}+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}\right]}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)

\(=\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}=\dfrac{30+6\sqrt{7}-18-6\sqrt{7}}{6\sqrt{2}}=\dfrac{12}{6\sqrt{2}}\)

\(=\sqrt{2}\)

\(x=\frac{\left(5+\sqrt{2}\right)^2\sqrt{\left(5-\sqrt{2}\right)^2}-\left(5-\sqrt{2}\right)^2\sqrt{\left(5+\sqrt{2}\right)^2}}{\frac{\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}-2\right)}+\sqrt{\left(\sqrt{13}+3\right)\left(\sqrt{13}-2\right)}}{\sqrt{13-4}}}\)

\(=\frac{\left(5+\sqrt{2}\right)\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)-\left(5-\sqrt{2}\right)\left(5-\sqrt{2}\right)\left(5+\sqrt{2}\right)}{\frac{\sqrt{19-5\sqrt{13}}+\sqrt{7+\sqrt{13}}}{3}}\)

\(=\frac{69\left(5+\sqrt{2}-5+\sqrt{2}\right)}{\frac{1}{\sqrt{2}}\left(\sqrt{38-10\sqrt{13}}+\sqrt{14+2\sqrt{13}}\right)}=\frac{276}{\sqrt{\left(5-\sqrt{13}\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}}\)

\(=\frac{276}{5-\sqrt{13}+\sqrt{13}+1}=46\)

\(\Rightarrow A=...\)