Rút gọn
\(\sqrt[3]{54+30\sqrt{3}}+\sqrt[3]{54-30\sqrt{3}}\)
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HT
14 tháng 10 2018
a) Đặt \(A=3+\sqrt{3}\)
<=>\(A^3=27+27\sqrt{3}+27+3\sqrt{3}\)
<=>\(A^3=54+30\sqrt{3}\)
<=>\(A=\sqrt[3]{54+30\sqrt{3}}\)
Vậy....
b) mình sửa lại đề nhá:
Tính \(B=\sqrt[3]{54+30\sqrt{3}}+\sqrt[3]{54-30\sqrt{3}}\)
\(B=\sqrt[3]{\left(3+\sqrt{3}\right)^3}+\sqrt[3]{\left(3-\sqrt{3}\right)^3}\)
\(B=3+\sqrt{3}+3-\sqrt{3}=6\)
TT
1
N
2
11 tháng 9 2021
\(5\sqrt[3]{2}+\sqrt[3]{-16}+\sqrt[3]{54}=5\sqrt[3]{2}-2\sqrt[3]{2}+3\sqrt[3]{2}=6\sqrt[3]{2}\)
11 tháng 9 2021
\(5\sqrt[3]{2}+\sqrt[3]{-16}+\sqrt[3]{54}\)
\(=5\sqrt[3]{2}-2\sqrt[3]{2}+3\sqrt[3]{2}\)
\(=6\sqrt[3]{2}\)
27 tháng 8 2018
a) \(\frac{x\sqrt[3]{y}+\sqrt[3]{x^2y^2}}{\sqrt[3]{x^2y^2}+y\sqrt[3]{x}}\)
\(=\frac{\sqrt[3]{x^2y}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)}{\sqrt[3]{xy^2}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)}=\sqrt[3]{\frac{x^2y}{xy^2}}=\sqrt[3]{\frac{x}{y}}\)
b) \(\frac{\sqrt[3]{54}-2\sqrt[3]{16}}{\sqrt[3]{54}+2\sqrt[3]{16}}\)
\(=\frac{\sqrt[3]{27.2}-2\sqrt[3]{8.2}}{\sqrt[3]{27.2}+2\sqrt[3]{8.2}}\)
\(=\frac{3\sqrt[3]{2}-4\sqrt[3]{2}}{3\sqrt[3]{2}+4\sqrt[3]{2}}=\frac{-\sqrt[3]{2}}{7\sqrt[3]{2}}=-\frac{1}{7}\)
DV
19 tháng 6 2021
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
5 tháng 4 2021
a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)
\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)
\(=5\sqrt{6}\)
5 tháng 4 2021
b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
Đặt \(A=\sqrt[3]{54+30\sqrt{3}}+\sqrt[3]{54-30\sqrt{3}}\)
\(=\sqrt[3]{27+27\sqrt{3}+3\sqrt{3}+27}+\sqrt[3]{27-27\sqrt{3}-3\sqrt{3}+27}\)
\(=\sqrt[3]{\left(3+\sqrt{3}\right)^3}+\sqrt[3]{\left(3-\sqrt{3}\right)^3}\)
\(=3+\sqrt{3}+3-\sqrt{3}\)
\(=6\)
Vậy \(A=6\)