Tính giá trị biểu thức :
\(M=\left(\frac{1}{4}\right)^{-\frac{3}{2}}-2\left(\frac{125}{27}\right)^{-\frac{2}{3}}+2^{\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}}\)
K
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Những câu hỏi liên quan
4 tháng 5 2016
\(A=\left(3\sqrt{3}\right)^{\frac{4}{3}}+\left(\frac{1}{16}\right)^{\frac{3}{4}}+2\left(\frac{8}{27}\right)^{\frac{2}{3}}\)
\(A=\left(3\sqrt{3}\right)^{\frac{4}{3}}+55+\frac{32}{3}\)
\(A=\left(3\sqrt{3}\right)^{\frac{4}{3}}+\frac{197}{3}\)
\(A=243+\frac{197}{3}\)
\(A=\frac{926}{3}\)
4 tháng 5 2016
Ta có \(A=3^{\frac{3}{2}.\frac{4}{3}}+\left(\frac{1}{2}\right)^{4.\frac{3}{4}}+2\left(\frac{2}{3}\right)^{3.\frac{2}{3}}=3^2+\left(\frac{1}{2}\right)^3+2\left(\frac{2}{3}\right)^2=\frac{721}{72}\)
17 tháng 9 2019
\(\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2+\left(4\sqrt{0,5}\right)^2-\left(\frac{1}{5}\sqrt{125}\right)^2\)
\(=2^2.3-3^2.2+4^2.0,5-5\)
\(=12-18+8-5\)
\(=-3\)
18 tháng 8 2023
a) \(\sqrt[4]{\dfrac{1}{16}}=\dfrac{1}{2}\)
b) \(\left(\sqrt[6]{8}\right)^2=\sqrt[\dfrac{6}{2}]{8}=\sqrt[3]{8}=2\)
c) \(\sqrt[4]{3}\cdot\sqrt[4]{27}=\sqrt[4]{3\cdot27}=\sqrt[4]{81}=3\)
KT
16 tháng 7 2018
\(B=\frac{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}{3+\sqrt{5}}=3-\sqrt{5}\)
\(C=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)
\(=\frac{-2\sqrt{3}}{2}=-\sqrt{3}\)
\(D=\frac{2}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}\)
\(=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{6\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+2\right)-\left(3-\sqrt{3}\right)\)
\(=\sqrt{3}-1-\sqrt{3}-2-3+\sqrt{3}=\sqrt{3}-6\)
Ta có:
\(\left(\frac{1}{4}\right)^{-\frac{3}{2}}=8\) ;
\(2\left(\frac{125}{27}\right)^{-\frac{2}{3}}=2.\frac{9}{25}=\frac{18}{25}\) ;
\(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}=2\Rightarrow2^{\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}}=2^2=4\)
\(\Rightarrow M=8-\frac{18}{25}+4=4\frac{18}{25}\)
Ta có \(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
Nên \(B=2^{2\left(-\frac{3}{2}\right)}-2\left(\frac{5}{3}\right)^{3\left(-\frac{2}{3}\right)}+2^2=2^3-2\left(\frac{3}{5}\right)^2+4=\frac{282}{25}\)