Đáp án:
Bài `5`
`a,`
`9/(2 × 5) + 9/(5 × 8) + 9/(8 × 11) + ... + 9/(29 × 32)`
`= 9 × [1/(2 × 5) + 1/(5×8) + 1/(8 × 11) + ... + 1/(29 × 32)]`
`= 9 × 1/3 × [1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/29 - 1/32]`
`= 3 × [1/2 + (- 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/29) - 1/32]`
`= 3 × [1/2 - 1/32]`
`= 3 × 15/32`
`= 45/32`
Ta thấy : `45/32 ≈ 1,4 > 1`
`-> 9/(2 × 5) + 9/(5 × 8) + 9/(8 × 11) + ... + 9/(29 × 32) > 1`
`b,`
Đặt `A = 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + ... + 1/60^2` `(1)`
Vì \(\left\{ \begin{array}{l}\dfrac{1}{3^2} = \dfrac{1}{9}\\ \dfrac{1}{4^2} < \dfrac{1}{3×4}\\ \dfrac{1}{5^2} < \dfrac{1}{4 × 5}\\ \dfrac{1}{6^2} < \dfrac{1}{5×6}\\.............\\ \dfrac{1}{60^2} < \dfrac{1}{59 × 60}\end{array} \right.\)
Nên từ `(1)`
`-> A < 1/9 + 1/(3 × 4) + 1/(4 × 5) + 1/(5 × 6) + ... + 1/(59 × 60)`
`-> A < 1/9 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/59 - 1/60`
`-> A < 1/9 + 1/3 + (- 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/59) - 1/60`
`-> A < 1/9 + 1/3 - 1/60`
`-> A < 4/9 - 1/60`
Ta thấy `4/9 - 1/60 < 4/9`
`-> A < 4/9 - 1/60 < 4/9`
`-> A < 4/9`
Bài `6`
`a,`
`[2/3x - x + (-1)/2x] ÷ 2/3 = -4`
`-> [(2/3 - 1 + (-1)/2) x] ÷ 2/3 = (-4)`
`-> (-5)/6x = -4 × 2/3`
`-> (-5)/6x = (-8)/3`
`-> x = (-8)/3 ÷ (-5)/6`
`-> x = 16/5`
Vậy `x = 16/5`
`b,`
`(x - 1)/2019 + (x - 2)/2020 = (x - 3)/2021 + (x - 4)/2022`
`-> (x - 1)/2019 + (x - 2)/2020 - (x - 3)/2021 - (x - 4)/2022 = 0`
`-> ( (x - 1)/2019 + 1) + ( (x - 2)/2020 + 1) - ( (x - 3)/2021 + 1) - ( (x - 4)/2022 + 1) = 0`
`-> ( (x - 1)/2019 + 2019/2019) + ( (x - 2)/2020 + 2020/2020) - ( (x - 3)/2021 + 2021/2021) - ( (x - 4)/2022 + 2022/2022) = 0`
`-> (x + 2018)/2019 + (x + 2018)/2020 - (x + 2018)/2021 - (x + 2018)/2022 = 0`
`-> (x + 2018) (1/2019 + 1/2020 - 1/2021 - 1/2022) = 0`
`-> x + 2018 = 0` (Vì `1/2019 + 1/2020 - 1/2021 - 1/2022 \ne 0`)
`-> x = -2018`
Vậy `x = -2018`
