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`1.`
`A= \root{3}{(4-2\sqrt 3)(\sqrt 3 -1)}`
Ta có
`4 - 2\sqrt 3 = 3 - 2\sqrt 3 +1 = (\sqrt 3-1)^2`
Khi đó
`A = \root{3}{(\sqrt 3-1)^2(\sqrt 3-1)}`
`A = \root{3}{(\sqrt 3-1)^3}`
`A = \sqrt 3-1`
`2`
`A = \root{3}{(\sqrt 2+1)(3+2\sqrt 2)}`
Ta có
`3+2\sqrt 2 = 2+2\sqrt 2 + 1 = (\sqrt 2+1)^2`
Khi đó
`A = \root{3}{(\sqrt 2+1)(\sqrt 2+1)^2}`
`A = \root{3}{(\sqrt 2+1)^3}`
`A = \sqrt 2+1`
`3`
`A = (\root{3}{4} + 1)^3 - (\root{3}{4} - 1)^3`
`A = (\root{3}{4} + 1 - \root{3}{4} + 2)[(\root{3}{4}+1)^2 + (\root{3}{4} +1)(\root{3}{4} - 1) + (\root{3}{4}-1)^2]`
`A = 2[\root{3}{16} + 2\root{3}{4} +1 + \root{3}{16} - 1 + \root{3}{16} - 2\root{3}{4} +1]`
`A = 2(3\root{3}{16} +1)`
`A = 6\root{3}{16} + 2`
`A = 12\root{3}{2} + 2`
`4`
`A= (\root{3}{9} -\root{3}{6} + \root{3}{4})(\root{3}{3} + \root {3}{2})`
`A = [(\root{3}{3})^2- \root{3}{3} . \root{3}{2} +(\root{3}{2})^2)(\root{3}{3} + \root{3}{2})`
`A = (\root{3}{3})^3 + (\root{3}{2})^2`
`A = 3 + 2`
`A= 5`
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$A = \sqrt[3]{(4 - 2\sqrt{3})(\sqrt{3} - 1)}$
$= \sqrt[3]{(3 - 2\sqrt{3} + 1)(\sqrt{3} - 1)}$
$= \sqrt[3]{(\sqrt{3} - 1)^2(\sqrt{3} - 1)}$
$= \sqrt[3]{(\sqrt{3} - 1)^3}$
$= \sqrt{3} - 1$
__________________________________
$A = \sqrt[3]{(\sqrt{2} + 1)(3 + 2\sqrt{2})}$
$= \sqrt[3]{(\sqrt{2} + 1)(2 + 2\sqrt{2} + 1)}$
$= \sqrt[3]{(\sqrt{2} + 1)(\sqrt{2} + 1)^2}$
$= \sqrt[3]{(\sqrt{2} + 1)^3}$
$= \sqrt{2} + 1$
__________________________________
$A = (\sqrt[3]{4} + 1)^3 - (\sqrt[3]{4} - 1)^3$
$= [(\sqrt[3]{4})^3 + 3(\sqrt[3]{4})^2 + 3\sqrt[3]{4} + 1] - [(\sqrt[3]{4})^3 - 3(\sqrt[3]{4})^2 + 3\sqrt[3]{4} - 1]$
$= 6(\sqrt[3]{4})^2 + 2$
$= 6\sqrt[3]{16} + 2$
$= 12\sqrt[3]{2} + 2$
__________________________________
$A = (\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4})(\sqrt[3]{3} + \sqrt[3]{2})$
$= (\sqrt[3]{3} + \sqrt[3]{2})[(\sqrt[3]{3})^2 - \sqrt[3]{3} \cdot \sqrt[3]{2} + (\sqrt[3]{2})^2]$
$= (\sqrt[3]{3})^3 + (\sqrt[3]{2})^3$
$= 3 + 2$
$= 5$
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