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30 tháng 5 2023
a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}\)
b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)
7 tháng 7 2021
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
24 tháng 4 2017
a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)
\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)
\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)
b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
12 tháng 6 2023
\(\sqrt{9-3\sqrt{8}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}+\sqrt{5-2\sqrt{6}}-\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2.\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
\(=\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{6}+\sqrt{2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{6}-\sqrt{3}\right|-\sqrt{6}+\sqrt{2}+\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{6}-\sqrt{3}-\sqrt{6}+\sqrt{2}+\sqrt{3}-\sqrt{2}\) (do \(\sqrt{6}-\sqrt{3}>0;\sqrt{3}-\sqrt{2}>0\))
\(=0\)
12 tháng 6 2023
\(=\sqrt{9-6\sqrt{2}}-\dfrac{\sqrt{6}-\sqrt{2}}{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{\sqrt{2}}\left(\sqrt{3}-1\right)\)
\(=\sqrt{6}-\sqrt{3}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}+\sqrt{3}-\sqrt{2}-\dfrac{1}{2}\sqrt{6}+\dfrac{1}{2}\sqrt{2}\)
\(=0\)
25 tháng 8 2021
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
VD
25 tháng 5 2022
`#Hưng`
\(a,3\sqrt{8\sqrt{5}}-2\sqrt{9\sqrt{20}}\\ =\sqrt{9.8\sqrt{5}}-\sqrt{4.9\sqrt{20}}\\ =\sqrt{72\sqrt{5}}-\sqrt{36\sqrt{20}}\\ =\sqrt{\sqrt{5184.5}}-\sqrt{\sqrt{1296.20}}\\ =\sqrt{\sqrt{25920}}-\sqrt{\sqrt{25920}}\\ =0\)
\(b,ĐKXĐ:x\sqrt{x}-\sqrt{x}+x-1\ne0\\ \Rightarrow\sqrt{x}\left(x-1\right)+\left(x-1\right)\ne0\\ \Rightarrow\left(x-1\right)\left(\sqrt{x}+1\right)\ne0\\ \Rightarrow x-1\ne0\left(vì.\sqrt{x}+1>0\right)\\ \Rightarrow x\ne1\)
24 tháng 6 2021
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
24 tháng 6 2021
`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
=\(16\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)=\(16\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)