Đáp án:
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Bài `5`
`a,`
`A = (-1)/20 + (-1)/30 + ... + (-1)/72 + (-1)/90`
`↔ A = -1/20 - 1/30 - ... - 1/72 - 1/90`
`↔ A = - [1/20 + 1/30 + ... + 1/72 + 1/90]`
`↔ A = - [1/(4×5) + 1/(5×6) + ... + 1/(8×9) + 1/(9×10) ]`
`↔ A = - [1/4 - 1/5 + 1/5 - 1/6 + ... + 1/8 - 1/9 + 1/9 - 1/10]`
`↔ A= - [1/4 + (-1/5 + 1/5) + (-1/6 + 1/6) + ... + (-1/9 + 1/9) - 1/10]`
`↔ A = - [1/4 - 1/10]`
`↔ A = (-3)/20`
Vậy `A = (-3)/20`
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`b,`
Có : `Q = (2010 + 2011 + 2012)/(2011 + 2012 + 2013)`
`↔ Q = 2010/(2011 + 2012 + 2013) + 2011/(2011 + 2012 + 2013) + 2012/(2011 +2012 + 2013)`
Có : \(\left\{ \begin{array}{l}2011 < 2011 + 2012 + 2013\\ 2012 < 2011+ 2012 + 2013\\2013 < 2011 + 2012 + 2013\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}\dfrac{1}{2011} > \dfrac{1}{2011 + 2012 + 2013}\\ \dfrac{1}{2012} > \dfrac{1}{2011+ 2012 + 2013}\\ \dfrac{1}{2013} > \dfrac{1}{2011 + 2012 + 2013}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}2010×\dfrac{1}{2011} > 2010 ×\dfrac{1}{2011 + 2012 + 2013}\\ 2011 ×\dfrac{1}{2012} > 2011 ×\dfrac{1}{2011+ 2012 + 2013}\\2012 × \dfrac{1}{2013} > 2012 ×\dfrac{1}{2011 + 2012 + 2013}\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}\dfrac{2010}{2011} > \dfrac{2010}{2011 + 2012 + 2013}\\ \dfrac{2011}{2012} > \dfrac{2011}{2011+ 2012 + 2013}\\\dfrac{2012}{2013} > \dfrac{2012}{2011 + 2012 + 2013}\end{array} \right.\)
Cộng theo vế ta được :
`↔ 2010/2011 + 2011/2012 + 2012/2013 > 2010/(2011 + 2012 + 2013) + 2011/(2011 + 2012 + 2013) + 2012/(2011 +2012 + 2013)`
`↔ P > Q`
Vậy `P > Q`
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Bài `6`
`A = (12n + 1)/(2n + 3)`
Để `A` nguyên
`↔ 12n + 1 \vdots 2n + 3`
`↔ 12n + 18- 17 \vdots 2n + 3`
`↔ 6 (2n + 3) - 17 \vdots 2n + 3`
Vì `6 (2n + 3) \vdots 2n+3`
`↔ -17 \vdots 2n + 3`
`↔2n + 3 ∈ Ư (-17) = {±1; ±17} (n ∈ ZZ)`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline 2n + 3& 1 & -1 & 17 & -17 \\\hline n& -1 & -2 & 7 & -10\\\hline\end{array}$
Vậy `n ∈ {-1;-2;7;-10}` để `A` nguyên
