(2/9)/3=(2/9)-3 (16/3)/4=(16/3)-4

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  • (2/9)/3=(2/9)-3,应改为:(9/2) /3=(9/2) -3;

    (16/3)/4=(16/3)-4

    (x/4)/5=(x/4)-5,(x/4)=5(x/4)-25,4(x/4)=25,x/4=25/4,x=25,∴ (25/4)/5=(25/4)-5,

    (9/2) /3=(9/2) -3;

    (16/3)/4=(16/3)-4

    (25/4)/5=(25/4)-5,

    (36/5)/6=(36/5)-6,

    (49/6)/7=(49/6)-7,

    (64/7)/8=(64/7)-8,

    .

    ∴[n²/(n-1)]/n=n²/(n-1)-n

    证明:左边=[n²/(n-1)]/n=n/(n-1)

    右边=n²/(n-1)-n=(n²-n²+n)/(n-1)=n/(n-1)

    左边=右边,等式成立.