Tính: a)\(\sqrt[3]{125}.\sqrt[3]{\dfrac{16}{10}}.\sqrt[3]{-0,5}\)b) \(\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)c) \(\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}\)d) \(\dfrac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}\)e)...
Đọc tiếp
Tính:
a)\(\sqrt[3]{125}.\sqrt[3]{\dfrac{16}{10}}.\sqrt[3]{-0,5}\)
b) \(\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
c) \(\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}\)
d) \(\dfrac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}\)
e) E=\(\sqrt[3]{2+10\sqrt{\dfrac{1}{27}}}+\sqrt[3]{2-10\sqrt{\dfrac{1}{27}}}\)
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 $E-2=0$
$\Leftrightarrow E=2$