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

Toán Lớp 9: Giúp e vs, hứa cho 5* và CTLHN :3 Thực hiện phép tính: $\sqrt{2-\sqrt{3}}$ – $\sqrt{2+\sqrt{3}}$

Toán Lớp 9: Giúp e vs, hứa cho 5* và CTLHN :3
Thực hiện phép tính:
$\sqrt{2-\sqrt{3}}$ – $\sqrt{2+\sqrt{3}}$

Comments ( 2 )

  1. \sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}
    =(\sqrt{4-2\sqrt{3}})/(\sqrt{2})-(\sqrt{4+2\sqrt{3}})/(\sqrt{2})
    =(\sqrt{(\sqrt{3}-1)^2})/(\sqrt{2})-(\sqrt{(\sqrt{3}+1)^2})/(\sqrt{2})
    =(|\sqrt{3}-1|)/(\sqrt{2})-(|\sqrt{3}+1|)/(\sqrt{2})
    =(\sqrt{3}-1)/(\sqrt{2})-(\sqrt{3}+1)/(\sqrt{2})
    =(\sqrt{3}-1-\sqrt{3}-1)/(\sqrt{2})
    =(-2)/(\sqrt{2})
    =-\sqrt{2}
     

  2. Đặt A =  $\sqrt{2-\sqrt{3}}$ – $\sqrt{2+\sqrt{3}}$
    ⇔ $\sqrt{2}$A = $\sqrt{4-2\sqrt{3}}$ – $\sqrt{4+2\sqrt{3}}$ 
    ⇔ $\sqrt{2}$A = $\sqrt{(\sqrt{3}-1)²}$ – $\sqrt{(\sqrt{3}+1)²}$
    ⇔ $\sqrt{2}$A = |$\sqrt{3}-1$|- |$\sqrt{3}+1$|
    ⇔ $\sqrt{2}$A = $\sqrt{3}-1$ – $\sqrt{3}-1$
    ⇔ $\sqrt{2}$A = -2
    ⇔ A = $\dfrac{-2}{\sqrt{2}}$ = $-\sqrt{2}$
    @Sano
    $\text{Chúc bn hok tốt}$
     

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