tính A=căn(4+2căn3) +căn(4-2căn3)

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tính
A=căn(4+2căn3) +căn(4-2căn3)

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Huy Gia 6 years 2020-10-15T06:01:35+00:00 2 Answers 208 views 0

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  1. dangkhoa
    0
    2020-10-15T06:03:03+00:00
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    $A=\sqrt[]{4+2\sqrt[]{3}}+\sqrt[]{4-2\sqrt[]{3}}$

    $=\sqrt[]{(\sqrt[]{3})^2+2\sqrt[]{3}.1+1^2}+\sqrt[]{(\sqrt[]{3})^2-2.\sqrt[]{3}.1+1^2}$

    $=\sqrt[]{(\sqrt[]{3}+1)^2}+\sqrt[]{(\sqrt[]{3}-1)^2}$

    $=|\sqrt[]{3}+1|+|\sqrt[]{3}-1|$

    $=\sqrt[]{3}+1+\sqrt[]{3}-1$ (vì √3-1>0)

    $=2\sqrt[]{3}$

     

  2. Euphemia
    0
    2020-10-15T06:03:15+00:00
    Reply

    √( 4 + 2√3 ) + √( 4 – 2√3 )

    = √( 3 +2√3 + 1 ) + √( 3 – 2√3 + 1 )

    = √( √3 + 1 )² + √( √3 – 1 )²

    = | √3 + 1 | + | √3 – 1 |

    = √3 + 1 + √3 – 1 

    = 2√3

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