= 5+(-7) - 2.4 - \(\dfrac{1}{3}\).6
= - 12
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Bài 9: Căn bậc ba
Các câu hỏi tương tự
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=\(\sqrt[3]{2+10\sqrt{\dfrac{1}{27}}}+\sqrt[3]{2-10\sqrt{\dfrac{1}{27}}}\)
tính
a. \(\dfrac{\sqrt[3]{125}.\sqrt[3]{\dfrac{16}{10}}.\sqrt[3]{-0,5}}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
b.\(\sqrt[]{3+\sqrt[]{5}+\sqrt[]{10+6\sqrt[]{5}}}\)
Tính
a.A=\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
b.B=\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{7}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{7}}}\)
c.C=\(\sqrt[3]{26+15\sqrt{3}}.\left(2-\sqrt{3}\right)+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)
Thực hiện các phép tính sau :
a)A=\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\) b)B=\(\left(2-\sqrt{3}\right).\sqrt[3]{26+15\sqrt{3}}\) c)C=\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{7}}}\)
c, Rút gọn.
a, \(\sqrt[3]{27a^3}-2a\) b, \(\sqrt[3]{27a^3}-\sqrt[3]{-8a^3}-\sqrt[3]{125a^3}\)
c, \(\sqrt[3]{16x^3}-\sqrt[3]{-54x^3}-\sqrt[3]{128x^3}\) d, \(\sqrt[3]{\dfrac{1}{8}y^6}+\sqrt[3]{\dfrac{1}{27}y^3}-\sqrt[3]{-\dfrac{1}{216}y^3}\)
H24
16 tháng 6 2021 lúc 21:48
Cho \(a=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\)
Chứng minh rằng: \(\dfrac{64}{\left(a^2-3\right)^3}-3a\) có giá trị là số nguyên
1.Tìm x:\(\left(x-3\right)^3\)=\(\dfrac{1}{64}\)
2.Chứng minh:
a,(\(\sqrt[3]{\sqrt[]{9+4\sqrt[]{5}}}\).\(\sqrt[3]{\sqrt[]{5.2}}\)).\(\sqrt[3]{\sqrt[]{5-2}}\) -2,1 <0
3.Rút gọn,\(\dfrac{\sqrt[3]{a^4}+\sqrt[3]{a^2b^2}+\sqrt[3]{b^4}}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}\)
giúp em với ạ tính
\(\sqrt[3]{125}.\sqrt[3]{\dfrac{16}{10}}.\sqrt[3]{-0,5}\)
1.Chứng minh:\(\dfrac{a+\sqrt{2+\sqrt{5}.}\sqrt{\sqrt{9-4\sqrt{5}}}}{3\sqrt{2-\sqrt{5}}.\sqrt[3]{\sqrt{9+4\sqrt{5}-}3\sqrt{a^2}+\sqrt[3]{a}}}\)=\(-\sqrt[3]{a}-1\)
2.Rút gọn: \(\left(\dfrac{a^3\sqrt[]{a}-2a^3\sqrt{b}+\sqrt[3]{a^2}-\sqrt[3]{b}}{\sqrt[3]{a^2-\sqrt[3]{ab}}}+\dfrac{\sqrt[3]{a^2b}-\sqrt[3]{ab^2}}{\sqrt[3]{a}-\sqrt[3]{b}}\right)1\dfrac{1}{\sqrt[3]{a^2}}\)