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Nguyễn Thanh Hằng CTV23 tháng 1 2018 lúc 17:19
4x+3⋮x−24x+3⋮x−2
Mà x−2⋮x−2x−2⋮x−2
⇔⎧⎨⎩4x+3⋮x−24x−8⋮x−2⇔{4x+3⋮x−24x−8⋮x−2
⇔11⋮x−2⇔11⋮x−2
⇔x−2∈Ư(11)⇔x−2∈Ư(11)
⇔⎡⎢ ⎢ ⎢⎣x−2=1x−2=11x−2=−1x−2=−11⇔[x−2=1x−2=11x−2=−1x−2=−11 ⇔⎡⎢ ⎢ ⎢⎣x=3x=13x=1x=−9⇔[x=3x=13x=1x=−9
Vậy ..
gọi (a,b) = d
Vì ab=45ab=45 mà (4,5) = 1⇒a=4d,b=5d⇒a=4d,b=5d
Ta có: ab=(a,b)⋅[a,b]⇒[a,b]=ab(a,b)=4d.5dd=4.5.d=20.d=140ab=(a,b)⋅[a,b]⇒[a,b]=ab(a,b)=4d.5dd=4.5.d=20.d=140
⇒d=7⇒{a=4.7=28b=5.7=35
a)3x-5=4x+6
=> 3x-4x=5+6
=> -1x=11
x=11:(-1)
x=-11
6 hình tứ giác
Ư (−2)={±1;±2}(−2)={±1;±2}
Ư (4)={±1;±2;±4}(4)={±1;±2;±4}
Ư (13)={±1;±13}(13)={±1;±13}
Ư (15)={±1;±3;±5;±15}(15)={±1;±3;±5;±15}
Ư (1)={±1}
n2+3n−13⋮n+3n2+3n−13⋮n+3
Mà n+3⋮n+3n+3⋮n+3
⇔⎧⎨⎩n2+3n−13⋮n+3n2+3n⋮n+3⇔{n2+3n−13⋮n+3n2+3n⋮n+3
⇔13⋮n+3⇔13⋮n+3
⇔n+3∈Ư(13)⇔n+3∈Ư(13)
⇔⎡⎢ ⎢ ⎢⎣n+3=1n+3=13n+3=−1n+3=−13⇔[n+3=1n+3=13n+3=−1n+3=−13
⇔⎡⎢ ⎢ ⎢⎣n=−2n=10n=−4n=−16⇔[n=−2n=10n=−4n=−16
a+b=c+d⇒a=c+d−ba+b=c+d⇒a=c+d−b
Ta có:ab+1=cdTa có:ab+1=cd
⇔(c+d−b)b+1=cd⇔(c+d−b)b+1=cd
⇔bc+bd−b2−cd=−1⇔bc+bd−b2−cd=−1
⇔c(b−d)−b(b−d)=−1⇔c(b−d)−b(b−d)=−1
⇔(b−d)(c−b)=−1⇔(b−d)(c−b)=−1
Vì b,c,d∈ZVì b,c,d∈Z
TH1:{b−d=1c−b=−1⇒{d=b−1c=b−1⇒c=dTH1:{b−d=1c−b=−1⇒{d=b−1c=b−1⇒c=d
TH2:{b−d=−1c−b=1⇒{d=b+1c=b+1⇒d=cTH2:{b−d=−1c−b=1⇒{d=b+1c=b+1⇒d=c
Vậy d=c