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a: \(=4\sqrt[3]{2}-9\sqrt[3]{2}++6\sqrt[3]{2}=\sqrt[3]{2}\)

b: \(=6\sqrt[3]{3}-15\sqrt[3]{3}+16\sqrt[3]{3}=7\sqrt[3]{3}\)

c: \(=-7\sqrt[3]{3}+3\sqrt[3]{3}+6\sqrt[3]{3}=2\sqrt[3]{3}\)

d: \(=8\sqrt[3]{5}-10\sqrt[3]{5}+2=-2\sqrt[3]{5}+2\)

b: Ta có: \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}\cdot\sqrt[3]{4}\)

\(=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54\cdot4}\)

=3-6

=-3

27 tháng 9 2023

`a)\root[3]{135}/\root[3]{5}-\root[3]{54}.\root[3]{4}`

`=\root[3]{135/5}-\root[3]{54.4}`

`=\root[3]{27}-\root[3]{216}`

`=3-6=-3`

`b)(\root[3]{25}-\root[3]{10}+\root[3]{4})(\root[3]{5}+\root[3]{2})`

`=5+\root[3]{50}-\root[3]{50}-\root[3]{20}+\root[3]{20}+2`

`=7`.

22 tháng 12 2023

a: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\dfrac{3}{2}\sqrt{2}+\dfrac{1}{2}\sqrt{2}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)

\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)

b: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{18}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)

c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

d: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=-\sqrt[3]{27}=-3\)

e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}=0\)

27 tháng 5 2021

\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)

27 tháng 5 2021

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).

12 tháng 10 2021

a: Ta có: \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)

\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)

=-5+2

=-3

12 tháng 10 2021

c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

8 tháng 7 2023

\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)

\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)

\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)

\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)

26 tháng 8 2021

a, \(\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)

\(=\left|\sqrt{2}-1\right|+\left|3\sqrt{2}-2\right|\)

\(=\sqrt{2}-1+3\sqrt{2}-2=4\sqrt{2}-3\)

b, \(2\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2\sqrt{3}+1\right)^2}\)

\(=2\left|\sqrt{3}-1\right|-\left|2\sqrt{3}+1\right|\)

\(=2\sqrt{3}-2-2\sqrt{3}-1=-3\)

26 tháng 8 2021

c, \(4\sqrt{\left(2-\dfrac{\sqrt{3}}{2}\right)^2}-3\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=4\left|2-\dfrac{\sqrt{3}}{2}\right|-3\left|\sqrt{3}-1\right|\)

\(=8-2\sqrt{3}-3\sqrt{3}+3=11-5\sqrt{3}\)