Tính:
a) \(\sqrt[3]{\left(\sqrt{2}+1\right)\left(3+2\sqrt{2}\right)}\)
b) \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
c) \(\left(\sqrt[3]{4}+1\right)^3-\left(\sqrt[3]{4}-1\right)^3\)
d) \(\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\)
e) \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
Mọi người giúp em với ạ!!!!!!!!!!!
a)
\(\sqrt[3]{(\sqrt{2}+1)(3+2\sqrt{2})}=\sqrt[3]{(\sqrt{2}+1)(2+2\sqrt{2}+1)}\)
\(=\sqrt[3]{(\sqrt{2}+1)(\sqrt{2}+1)^2}=\sqrt[3]{(\sqrt{2}+1)^3}=\sqrt{2}+1\)
b)
\(\sqrt[3]{(4-2\sqrt{3})(\sqrt{3}-1)}=\sqrt[3]{(3-2\sqrt{3}+1)(\sqrt{3}-1)}\)
\(=\sqrt[3]{(\sqrt{3}-1)^2(\sqrt{3}-1)}=\sqrt[3]{(\sqrt{3}-1)^3}=\sqrt{3}-1\)
c)
\((\sqrt[3]{4}+1)^3-(\sqrt[3]{4}-1)^3=[(\sqrt[3]{4}+1-(\sqrt[3]{4}-1)][(\sqrt[3]{4}+1)^2+(\sqrt[3]{4}+1)(\sqrt[3]{4}-1)+(\sqrt[3]{4}-1)^2]\)
\(=2[\sqrt[3]{16}+1+2\sqrt[3]{4}+\sqrt[3]{16}-1+\sqrt[3]{16}+1-2\sqrt[3]{4}]\)
\(=2(3\sqrt[3]{16}+1)\)
d)
\((\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}+\sqrt[3]{2})=[(\sqrt[3]{3})^2-\sqrt[3]{3}.\sqrt[3]{2}+(\sqrt[3]{2})^2](\sqrt[3]{3}+\sqrt[3]{2})\)
\(=(\sqrt[3]{3})^3+(\sqrt[3]{2})^3=3+2=5\)
e)
\(E=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
Áp dụng công thức $(a+b)^3=a^3+b^3+3ab(a+b)$ ta có:
\(E^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{(20+14\sqrt{2})(20-14\sqrt{2})}.E\)
\(E^3=40+3\sqrt[3]{20^2-(14\sqrt{2})^2}.E\)
\(E^3=40+3\sqrt[3]{8}.E=40+6E\)
\(\Leftrightarrow E^2(E-4)+4E(E-4)+10(E-4)=0\)
\(\Leftrightarrow (E-4)(E^2+4E+10)=0\)
Dễ thấy $E^2+4E+10=(E+2)^2+6\neq 0$ nên $E-4=0$ hay $E=4$
