Đáp án:
Giải thích các bước giải:
$B=\dfrac{5}{4}+\dfrac{10}{9}+...+\dfrac{2501}{2500}$
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$=\dfrac{4+1}{4}+\dfrac{9+1}{9}+...+\dfrac{2500+1}{2500}$
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$=1+\dfrac{1}{4}+1+\dfrac{1}{9}+...+1+\dfrac{1}{2500}$
$ $
$=(1+1+...+1)+(\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+...+\dfrac{1}{50^{2}})$
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Ta có: $\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+...+\dfrac{1}{50^{2}}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}$
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$⇒\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+...+\dfrac{1}{50^{2}}<1-\dfrac{1}{50}$
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$⇒\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+...+\dfrac{1}{50^{2}}<1$
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$⇒(\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+...+\dfrac{1}{50^{2}})∉Z$
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Mà $(1+1+...+1)∈Z$
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$⇒(1+1+...+1)+(\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+...+\dfrac{1}{50^{2}})∉Z$
$ $
$⇒B∉Z$
