Bài 5:
$\text{Gọi d là ƯCLN(2n+5;3n+7)}$
Vì `n in ZZ=>2n+5;3n+7 inZZ=>d inZZ`
`=>`$\left \{ {{2n+5 ⋮d} \atop {3n+7⋮d}} \right.$ `=>`$\left \{ {{3(2n+5) ⋮d} \atop {2(3n+7)⋮d}} \right.$`=>`$\left \{ {{6n+15 ⋮d} \atop {6n+14⋮d}} \right.$
`=>(6n+15)-(6n+14)⋮d`
`=>1⋮d`
`=>d in Ư(1)={+-1}`
`=>ƯCLN(2n+5;3n+7)={+-1}`
`=>[2n+5]/[3n+7]` là phân số tối giản.`(n in ZZ)`
Vậy `[2n+5]/[3n+7]` là phân số tối giản.`(n in ZZ)`
Bài 6:
*`A=(1/2^2-1)(1/3^2-1)(1/4^2-1)...(1/21^2-1)`
`A=([1^2-2^2]/2^2)([1^2-3^2]/3^2)([1^2-4^2]/4^2)...([1^2-21^2]/21^2)`
`A=[(1-2)(1+2)(1-3)(1+3)(1-4)(1+4)...(1-21)(1+21)]/[2^2. 3^2. 4^2...21^2]`
`A=[(-1).(-2).(-3)...(-20).3.4.5...22]/[2.2.3.3.4.4...21.21]`
`A=-[(1.2.3...20).(3.4.5...22)]/[(2.3.4...21).(2.3.4...21)]`
`A=-[22]/[21.2]`
`A=-11/21`
*`B=(1/2-1)(1/3-1)(1/4-1)...(1/2021-1)`
`B=(1/2-2/2)(1/3-3/3)(1/4-4/4)...(1/2021-2021/2021)`
`B=-1/2.-2/3.-3/4...-2020/2021`
`B=[(-1).(-2).(-3)...(-2020)]/[2.3.4...2021]`
Vì từ `1 to2020` có chẵn chữ số nên `(-1).(-2).(-3)...(-2020)=1.2.3...2020`
`=>B=[1.2.3...2020]/[2.3.4...2021]`
`B=1/2021`
$#Rabifoot$
