Đáp án:
`a,`
`A = 2/(1×5) + 3/(5×11) + 4/(11×19) + 5/(19×29) + 6/(29×31)`
Nhân cả 2 vế cho `2` ta được :
`⇔ 2A = 2 × ( 2/(1×5) + 3/(5×11) + 4/(11×19) + 5/(19×29) + 6/(29×31) )`
`⇔ 2A = 4/(1 × 5) + 6/(5 × 11) + 8/(11 × 19) + 10/(19 × 29) + 12/(29 × 31)` `(1)`
Đặt `H = 4/(1 × 5) + 6/(5 ×11) + 8/(11 × 19) + 10/(19 × 29)`
`⇔ H = 1 - 1/5 + 1/5 - 1/11 + 1/11 - 1/19 + 1/19 - 1/29`
`⇔ H = 1 + ( - 1/5 + 1/5 - 1/11 + 1/11 - 1/19 + 1/19) - 1/29`
`-> H = 1 - 1/29`
`-> H = 28/29`
Thay vào `(1)` ta được :
`⇔ 2A = 28/29 + 12/(29 × 31)`
`⇔ A = (28/29 + 12/(29 × 31) ) ÷ 2`
`⇔ A = 28/29 × 1/2 + 12/(29 × 31) × 1/2`
`⇔ A = 14/29 + 6/(29 × 31)`
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`B = 1/(1 × 4) + 2/(4 × 10) + 3/(10 × 19) + 4/(19 × 31)`
Nhân cả 2 vế cho `3` ta được :
`⇔ 3B = 3 × (1/(1 × 4) + 2/(4 × 10) + 3/(10 × 19) + 4/(19 × 31) )`
`⇔ 3B = 3/(1 × 4) + 6/(4 × 10) + 9/(10 × 19) + 12/(19 × 31)`
`⇔ 3B = 1 - 1/4 + 1/4 - 1/10 + 1/10 - 1/19 + 1/19 - 1/31`
`⇔ 3B = 1 + (- 1/4 + 1/4 - 1/10 + 1/10 - 1/19 + 1/19) - 1/31`
`⇔ 3B = 1 - 1/31`
`⇔ B = (1 - 1/31) ÷ 3`
`⇔ B = 1 × 1/3 - 1/31 × 1/3`
`⇔ B = 1/3 - 1/93`
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Có : \(\left\{ \begin{array}{l}A = \dfrac{14}{29} + \dfrac{6}{29×31}\\B = \dfrac{1}{3} - \dfrac{1}{93}\end{array} \right.\)
Ta thấy : `B = 1/3 - 1/93 < 1/3`
`-> B < 1/3`
Có : `14/29 = 42/87, 1/3 = 29/87`
`-> 42/87 > 29/87`
`-> 14/29 > 1/3`
`-> 14/29 + 6/(29 × 31) > 1/3`
`-> A > 1/3 > B`
`-> A > B`
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$b,$
`S = 1/2020 - 5/(2 × 4) - 5/(4 × 6) - 5/(6 × 8) - ... - 5/(2020 × 2022)`
`⇔ S = 1/2020 - (5/(2×4) + 5/(4×6) + 5/(6 × 8) + ... + 5/(2020 × 2022) )` `(1)`
Đặt `A = 5/(2×4) + 5/(4×6) + 5/(6 × 8) + ... + 5/(2020 × 2022)`
`⇔ A = 5 × [1/(2 × 4) + 1/(4 × 6) + 1/(6 × 8) + ... + 1/(2020 × 2022) ]`
`⇔ A = 5 × 1/2 × [1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2020 - 1/2022]`
`⇔ A =5/2 × [1/2 - 1/2022]`
`⇔ A = 5/2 × 505/1011`
`⇔ A = 2525/2022`
Thay vào `(1)` ta được :
`⇔ S = 1/2020 - 2525/2022`
Vậy `S = 1/2020 - 2525/2022`
