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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Rút gọn biểu thức: a) √2(√(2-√3) )(√3+1) b) √(2-√3) (√6-√2)(2+√3) Giải chi tiết giùm mình nhé ( cảm ơn nhiều ^^)

Toán Lớp 9: Rút gọn biểu thức:
a) √2(√(2-√3) )(√3+1)
b) √(2-√3) (√6-√2)(2+√3)
Giải chi tiết giùm mình nhé ( cảm ơn nhiều ^^)

Comments ( 1 )

  1. Giải đáp:
    $a. \sqrt[]{2}.\sqrt[]{2-\sqrt[]{3}}.( \sqrt[]{3} + 1 )$
    $= \sqrt[]{4-2\sqrt[]{3}}.( \sqrt[]{3} + 1 )$
    $= \sqrt[]{3-2\sqrt[]{3}+1}.( \sqrt[]{3} + 1 )$
    $= \sqrt[]{(\sqrt[]{3}-1)^{2}} . ( \sqrt[]{3} + 1 )$
    $= | \sqrt[]{3} – 1 |.( \sqrt[]{3} + 1 )$
    $= ( \sqrt[]{3} – 1 )( \sqrt[]{3} + 1 )$
    $= 3 – 1$
    $= 2$
    ( vì $\sqrt[]{3} – 1 > 0 ⇒ | \sqrt[]{3} – 1 | = \sqrt[]{3} – 1$ )
    $b. \sqrt[]{2-\sqrt[]{3}}.( \sqrt[]{6} – \sqrt[]{2} )( 2 + \sqrt[]{3} )$
    $= \sqrt[]{4-2\sqrt[]{3}}.( \sqrt[]{3} – 1 )( 2 + \sqrt[]{3} )$
    $= \sqrt[]{3-2\sqrt[]{3}+1}.( \sqrt[]{3} – 1 )( 2 + \sqrt[]{3} )$
    $= \sqrt[]{(\sqrt[]{3}-1)^{2}}.( \sqrt[]{3} – 1 )( 2 + \sqrt[]{3} )$
    $= | \sqrt[]{3} – 1 |.( \sqrt[]{3} – 1 )( 2 + \sqrt[]{3} )$
    $= ( \sqrt[]{3} – 1 )( \sqrt[]{3} – 1 )( 2 + \sqrt[]{3} )$
    $= ( 3 – 2\sqrt[]{3} + 1 )( 2 + \sqrt[]{3} )$
    $= ( 4 – 2\sqrt[]{3} )( 2 + \sqrt[]{3} )$
    $= 2( 2 – \sqrt[]{3} )( 2 + \sqrt[]{3} )$
    $= 2( 4 – 3 )$
    $= 2$
    ( vì $\sqrt[]{3} – 1 > 0 ⇒ | \sqrt[]{3} – 1 | = \sqrt[]{3} – 1$ )

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222-9+11+12:2*14+14 = ? ( )

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