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

Toán Lớp 9: 1/1+√2 +1/√2+√3+…+1/√99+√100

Toán Lớp 9: 1/1+√2 +1/√2+√3+…+1/√99+√100

Comments ( 2 )

  1. Giải đáp:
    $9$ 
    Lời giải và giải thích chi tiết:
    Ta có: $\dfrac{1}{\sqrt{n}+\sqrt{n+1}}$
    $=\dfrac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n+1}-\sqrt{n})}$
    $=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}$
    $=\sqrt{n+1}-\sqrt{n}$
    Do đó biểu thức trở thành
    $\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+…+\dfrac{1}{\sqrt{99}+\sqrt{100}}$
    $=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+…+\sqrt{100}-\sqrt{99}$
    $=\sqrt{100}-1$
    $=9$

  2. \qquad 1/(1+\sqrt{2})+1/(\sqrt{2}+\sqrt{3})+….+1/(\sqrt{99}+\sqrt{100})
    =(1-\sqrt{2})/(1-2)+(\sqrt{2}-\sqrt{3})/(2-3)+….+(\sqrt{99}-\sqrt{100})/(99-100)
    =sqrt{2}-1+sqrt{3}-sqrt{2}+….+sqrt{100}-sqrt{99}
    =\sqrt{100}-1
    =10-1
    =9
     

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

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