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

Toán Lớp 6: so sánh E=n/n+1 : F=n+2 /n+3

Toán Lớp 6: so sánh
E=n/n+1 : F=n+2 /n+3

Comments ( 2 )

  1. Giải đáp:
    $\rm E<F$
    Lời giải và giải thích chi tiết:
    $\rm E=\dfrac{n}{n+1}\\\Rightarrow E=\dfrac{n+1-1}{n+1}\\\Rightarrow E=\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\\\Rightarrow E=1-\dfrac{1}{n+1}\\F=\dfrac{n+2}{n+3}\\\Rightarrow F=\dfrac{n+3-1}{n+3}\\\Rightarrow F=\dfrac{n+3}{n+3}-\dfrac{1}{n+3}\\\Rightarrow F=1-\dfrac{1}{n+3}\\\Rightarrow\dfrac{1}{n+1}>\dfrac{1}{n+3}\\\Rightarrow1-\dfrac{1}{n+1}<1-\dfrac{1}{n+3}\\\Rightarrow\dfrac{n}{n+1}<\dfrac{n+2}{n+3}$
    Vậy $\rm E<P$.

  2. Giải đáp:
    E<F
    Lời giải và giải thích chi tiết:
    Ta có:
    Vì E<1=>E<[n+2]/[n+1+2]=[n+2]/[n+3]=F=>E<F
    Vậy E<F

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

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