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Gọi \(d=ƯC\left(a,b\right)\)
\(\Rightarrow\hept{\begin{cases}a⋮d\\b⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(21n+1\right)⋮d\\\left(14n+3\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2\left(21n+1\right)⋮d\\3\left(14n+3\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(42n+2\right)⋮d\\\left(42n+9\right)⋮d\end{cases}}\)
\(\Rightarrow\left[\left(42n+9\right)-\left(42n+2\right)\right]⋮d\)
\(\Rightarrow7⋮d\)
\(\RightarrowƯC\left(a,b\right)=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
Ta thấy trong các ước của 7 thì ước 7 là ước lớn nhất
Vậy \(ƯCLN\left(a,b\right)=7\)
Gọi d là ước chung của n+3 và 2n+5
Ta có : n+3 chia hết cho d
Suy ra (2n+6) - ( 2n+5) chia hết cho d => 1 chia hết cho d.
Vây d = 1
a)Gọi ƯCLN (\(n+3;2n+5\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(n+3\right)⋮d\Rightarrow2\left(n+3\right)⋮d\Rightarrow\left(2n+6\right)⋮d\\\left(2n+5\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN (\(n+3;2n+5\))=1
\(\Rightarrow\frac{n+3}{2n+5}\)là phân số tối giản(đpcm)
b)Gọi ƯCLN (\(2n+9;3n+14\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(2n+9\right)⋮d\Rightarrow3\left(2n+9\right)⋮d\Rightarrow\left(6n+27\right)⋮d\\\left(3n+14\right)⋮d\Rightarrow2\left(3n+14\right)⋮d\Rightarrow\left(6n+28\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+28\right)-\left(6n+27\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN (\(2n+9;3n+14\))=1
\(\Rightarrow\frac{2n+9}{3n+14}\) là phân số tối giản.(đpcm)
c)Gọi ƯCLN(\(6n+11;2n+5\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(6n+11\right)⋮d\\\left(2n+5\right)⋮d\Rightarrow3\left(2n+5\right)⋮d\Rightarrow\left(6n+15\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+15\right)-\left(6n+11\right)⋮d\)
\(\Rightarrow4⋮d\)
Mà \(\left(6n+15\right);\left(6n+11\right)⋮̸2\)
\(\Rightarrow d=1\)
⇒ƯCLN(\(6n+11;2n+5\))=1
\(\Rightarrow\frac{6n+11}{2n+5}\)là phân số tối giản (đpcm)
d)Gọi ƯCLN(\(12n+1;30n+2\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(12n+1\right)⋮d\Rightarrow5\left(12n+1\right)⋮d\Rightarrow\left(60n+5\right)⋮d\\\left(30n+2\right)⋮d\Rightarrow2\left(30n+2\right)⋮d\Rightarrow\left(60n+4\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(60n+5\right)-\left(60n+4\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(12n+1;30n+2\))=1
\(\Rightarrow\frac{12n+1}{30n+2}\) là phân số tối giản (đpcm)
e)Gọi ƯCLN(\(21n+4;14n+3\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(21n+4\right)⋮d\Rightarrow2\left(21n+4\right)⋮d\Rightarrow\left(42n+8\right)⋮d\\\left(14n+3\right)⋮d\Rightarrow3\left(14n+3\right)⋮d\Rightarrow\left(42n+9\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(42n+9\right)-\left(42n+8\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(21n+4;14n+3\))=1
\(\Rightarrow\frac{21n+4}{14n+3}\)là phân số tối giản (đpcm)
f) Gọi ƯCLN(\(2n+3;n+2\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(2n+3\right)⋮d\\\left(n+2\right)⋮d\Rightarrow2\left(n+2\right)⋮d\Rightarrow\left(2n+4\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+4\right)-\left(2n+3\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(2n+3;n+2\))=1
\(\Rightarrow\frac{2n+3}{n+2}\)là phân số tối giản (đpcm)
g) Gọi ƯCLN(\(n+1;3n+2\))=d
\(\Rightarrow\left\{{}\begin{matrix}\left(n+1\right)⋮d\Rightarrow3\left(n+1\right)⋮d\Rightarrow\left(3n+3\right)⋮d\\\left(3n+2\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left(3n+3\right)-\left(3n+2\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)
⇒ƯCLN(\(n+1;3n+2\))=1
\(\Rightarrow\frac{n+1}{3n+2}\) là phân số tối giản (đpcm)
a) Gọi d là ƯCLN(n, n + 1), d ∈ N*
\(\Rightarrow\hept{\begin{cases}n⋮d\\n+1⋮d\end{cases}}\)
\(\Rightarrow\left(n+1\right)-n⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(n,n+1\right)=1\)
\(\Rightarrow\) \(\frac{n}{n+1}\) là phân số tối giản.
b) Gọi d là ƯCLN(n + 1, 2n + 3), d ∈ N*
\(\Rightarrow\hept{\begin{cases}n+1⋮d\\2n+3⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(n+1\right)⋮d\\2n+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}2n+2⋮d\\2n+3⋮d\end{cases}}}\)
\(\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(n+1,2n+3\right)=1\)
\(\Rightarrow\) \(\frac{n+1}{2n+3}\) là phân số tối giản.
c) Gọi d là ƯCLN(21n + 4, 14n + 3), d ∈ N*
\(\Rightarrow\hept{\begin{cases}21n+4⋮d\\14n+3⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(21n+4\right)⋮d\\3\left(14n+3\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}42n+8⋮d\\42n+9⋮d\end{cases}}}\)
\(\Rightarrow\left(42n+9\right)-\left(42n+8\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(21n+4,14n+3\right)=1\)
\(\Rightarrow\) \(\frac{21n+4}{14n+3}\) là phân số tối giản.
d) Gọi d là ƯCLN(2n + 3, 3n + 5), d ∈ N*
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\3n+5⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+3\right)⋮d\\2\left(3n+5\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}6n+9⋮d\\6n+10⋮d\end{cases}}}\)
\(\Rightarrow\left(6n+10\right)-\left(6n+9\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(2n+3,3n+5\right)=1\)
\(\Rightarrow\) \(\frac{2n+3}{3n+5}\) là phân số tối giản.
b,
b.a=30=1.30=2.15=3.10=5.6
=>(b,a)={(1,30),(2,15),(3,10),(5,6)}
c,
(x+1)(y+2)=10=1.10=2.5
TH1:x+1=1;y+2=10=>x=0,y=8
tuong tu=>(x,y)={(0,8),(1,3),(4,0)}
\(TH1;n=3k\)\(\Rightarrow10^n+18n-1=\)\(10^{3k}+18.3k-1=1000^k+54k-1\equiv1+54k-1\left(mod27\right)\equiv0\left(mod27\right)\left(1\right)\)
\(TH2;n=3k+1\Rightarrow10^n+18n-1=10^{3k+1}+18.\left(3k+1\right)-1\)\(=10^{3k}.10+18.\left(3k+1\right)-1=1000^k.10+54k+18-1\)\(\equiv1.10+54k+17\left(mod27\right)\equiv54k+27\left(mod27\right)\equiv0\left(mod27\right)\left(2\right)\)
\(TH3;n=3k+2\Rightarrow10^n+18n-1=10^{3k+2}+54k+36-1\)\(=1000^{3k}.100+54k+35\equiv1.100+54k+35\left(mod27\right)\)\(\equiv54k+135\left(mod27\right)\equiv0\left(mod27\right)\left(3\right)\)\(Từ\left(1\right);\left(2\right);\left(3\right)\Rightarrow10^n+18n-1⋮27,\forall n\in N\left(ĐPCM\right)\)
Có :\(n-6⋮n-1\)
\(\Rightarrow n-1-5⋮n-1\)
Để n - 6 chia hết cho n-1
\(\Rightarrow5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(5\right)=\left(1;-1;5;-5\right)\)
\(\Rightarrow n\in\left(2;0;6;-4\right)\)
Gọi UC(a;b)=d
=>a=21n+1 chia hết cho d
b=14n+3 chia hết cho d
=>2(21n+1) chia hết cho d
3(14n+3) chia hết cho d
Hay 42n+2 chia hết cho d
42n+9 chia hết cho d
=>(42n+9)-(42n+2) chia hết cho d
=>7 chia hết cho d
=>d thuộc Ư(7)=(-7;-1;7;1)
Vậy UC(a;b)=(-7;-1;7;1)
~~~Xin lỗi bạn vì mình không ghi được dấu ngoặc nhọn và dấu chia hết!!! Sorry~~~