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

Toán Lớp 9: (3+2√3 / √3+2 + 2+ √2 / √2+1):(1 : 1 / √2+ √3)=1

Toán Lớp 9: (3+2√3 / √3+2 + 2+ √2 / √2+1):(1 : 1 / √2+ √3)=1

Comments ( 2 )

  1. $(\dfrac{3+2\sqrt{3}}{\sqrt{3}+2}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}):(1:\dfrac{1}{\sqrt{2}+\sqrt{3}})$
    $=(\dfrac{\sqrt{3}(\sqrt{3}+2)}{\sqrt{3}+2}+\dfrac{\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}+1}:(1:\dfrac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})})$
    $=(\sqrt{3}+\sqrt{2}):(1.\dfrac{-1}{\sqrt{2}-\sqrt{3}})$
    $=(\sqrt{3}+\sqrt{2}).\dfrac{-1}{\sqrt{2}-\sqrt{3}}$
    $=1$
    Vậy $(\dfrac{3+2\sqrt{3}}{\sqrt{3}+2}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}):(1:\dfrac{1}{\sqrt{2}+\sqrt{3}})=1$(đpcm)
     

  2. Giải đáp:
     Chứng minh đẳng thức:((3+2\sqrt3)/(\sqrt3+2)+(2+\sqrt2)/(\sqrt2+1)):(1:1/(\sqrt2+\sqrt3))=1
    Lời giải và giải thích chi tiết:
    ((3+2\sqrt3)/(\sqrt3+2)+(2+\sqrt2)/(\sqrt2+1)):(1:1/(\sqrt2+\sqrt3))
    =((\sqrt3(\sqrt3+2))/(\sqrt3+2)+(\sqrt2(\sqrt2+1))/\(\sqrt2+1)):((\sqrt2+\sqrt3)/1)
    =(\sqrt3+\sqrt2):(\sqrt3+\sqrt2)
    =1

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

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