\(\sqrt{72}+\sqrt{4\frac{1}{2}}\)
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MM
1
20 tháng 7 2020
Ta có: \(A=\sqrt{72}-6\sqrt{5\frac{1}{3}}+4\sqrt{12\frac{1}{2}}+2\sqrt{27}\)
\(=\sqrt{72}-\sqrt{36\cdot\frac{16}{3}}+\sqrt{16\cdot\frac{25}{2}}+\sqrt{108}\)
\(=\sqrt{72}-\sqrt{192}+\sqrt{200}+\sqrt{108}\)
\(=\left(\sqrt{72}+\sqrt{200}\right)-\left(\sqrt{192}-\sqrt{108}\right)\)
\(=6\sqrt{2}+10\sqrt{2}-\left(8\sqrt{3}-6\sqrt{3}\right)\)
\(=16\sqrt{2}-2\sqrt{3}\)
T
21 tháng 10 2019
Có
\(\frac{1}{2}\sqrt{72}+\frac{3}{4}\sqrt{48}+\sqrt{162}-\sqrt{75}=3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}=12\sqrt{2}-2\sqrt{3}\)
\(\sqrt[3]{125}+\sqrt[3]{-343}-2\sqrt[3]{64}+\frac{1}{3}\sqrt[3]{126}=5-7-8+\frac{1}{3}\sqrt[3]{126}=\frac{1}{3}\sqrt[3]{126}-10\)
ND
21 tháng 10 2019
a, = \(3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)
= \(12\sqrt{2}-2\sqrt{3}\)
b, = 5 - 7 - 8 + 2
= - 8
14 tháng 10 2020
a) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)
\(=\frac{3}{2}\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)
\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)
b) \(=\frac{\sqrt{5}+3-\sqrt{5}+3}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}.\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\)
\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)
c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)
\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)
TH
1
1 tháng 8 2016
\(\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}-\frac{1+\sqrt{2}}{1-\sqrt{2}}\right):\sqrt{72}\)
\(\left[\frac{\left(1-\sqrt{2}\right)^2-\left(1+\sqrt{2}\right)^2}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\right]:\sqrt{72}\)
\(=\frac{1-2\sqrt{2}+2-1-2\sqrt{2}-2}{1-2}\cdot\frac{1}{\sqrt{72}}\)
\(=\frac{-2\sqrt{2}-2\sqrt{2}}{-1}\cdot\frac{1}{\sqrt{72}}\)
\(=4\sqrt{2}\cdot\frac{1}{2\sqrt{18}}=\frac{2}{\sqrt{9}}=\frac{2}{3}\)
\(\sqrt{72}+\sqrt{4\frac{1}{2}}\)= \(\sqrt{72}+\sqrt{\frac{9}{2}}\)=\(\sqrt{72}+\frac{\sqrt{9}}{\sqrt{2}}\)=\(\sqrt{72}+\frac{3}{\sqrt{2}}\)
= \(\sqrt{36.2}+\frac{3}{\sqrt{2}}\)= \(6\sqrt{2}+\frac{3}{\sqrt{2}}\)=\(\frac{6.\left(\sqrt{2}\right)^2}{\sqrt{2}}+\frac{3}{\sqrt{2}}\)
= \(\frac{15}{\sqrt{2}}\)
\(\sqrt{72}+\sqrt{4\frac{1}{2}}\)
\(=6\sqrt{2}+\frac{3}{2}\sqrt{2}\)
\(=\left(6+\frac{3}{2}\right)\sqrt{2}\)
\(=\frac{15}{2}\sqrt{2}\)