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Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
Gọi Ư(n+1;2n+3) = d ( \(d\in\)N*)
\(n+1=2n+2\left(1\right);2n+3\left(2\right)\)
Lấy (2 ) - (1) ta được : \(2n+3-2n+2=1⋮d\Rightarrow d=1\)
Vậy ta có đpcm
Gọi Ư\(\left(3n+2;5n+3\right)=d\)( d \(\in\)N*)
\(3n+2=15n+10\left(1\right);5n+3=15n+9\left(2\right)\)
Lấy (!) - (2) ta được : \(15n+10-15n-9=1⋮d\Rightarrow d=1\)
Vậy ta có đpcm
a) Gọi \(d\) là UCLN \(\left(n+1,2n+3\right)\left(d\in N\right)\)
Ta có : \(\left[{}\begin{matrix}n+1⋮d\\2n+3⋮d\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2n+2⋮d\\2n+3⋮d\end{matrix}\right.\)
\(\Rightarrow2n+3-\left(2n+2\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\left(đpcm\right)\)
b) Gọi \(d\) là \(UCLN\left(2n+3,4n+8\right)\left(d\in N\right)\)
Ta có : \(\left[{}\begin{matrix}2n+3⋮d\\4n+8⋮d\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4n+6⋮d\\4n+8⋮d\end{matrix}\right.\)
\(\Rightarrow4n+8-\left(4n+6\right)⋮d\)
\(\Rightarrow2⋮d\)
\(\Rightarrow d\in\left\{1;2\right\}\)
Mà 2n+3 là số lẻ nên
\(\Rightarrow d=1\left(đpcm\right)\)
c) Gọi \(d\) là \(UCLN\left(3n+2;5n+3\right)\left(d\in N\right)\)
Ta có : \(\left[{}\begin{matrix}3n+2⋮d\\5n+3⋮d\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}15n+10⋮d\\15n+9⋮d\end{matrix}\right.\)
\(\Rightarrow15n+10-\left(15n+9\right)⋮d\)
\(\Rightarrow d=1\left(đpcm\right)\)
Gọi Ư( n+1; 2 n+3 ) = d ( d∈N* )
n +1 = 2n + 2 (1) ; 2n+3*) (2)
Lấy (2 ) - (1) ta được : 2n + 3 - 2n + 2 = 1:d => d =1
vậy ta có đpcm
gọi Ư ( 3n + 2 ; 5n + 3 ) = d ( d∈N* )
3n +2 = 15 n + 10 (1) ; 5n + 3 =15n + 9 (2)
lấy (!) - (2) ta được 15n + 10 - 15n - 9 = 1:d => d = 1
Vậy ta có đpcm
b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)
=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)
=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101
=>1-1/(n+1)=100/101
=>1/(n+1)=1/101
=>n+1=101
=>n=100
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
a: Gọi d=ƯCLN(15n+1;30n+1)
=>30n+2-30n-1 chia hết cho d
=>1 chia hết cho d
=>Đây là phân số tối giản
b: Gọi d=ƯCLN(3n+2;5n+3)
=>15n+10-15n-9 chia hết cho d
=>1 chia hết cho d
=>d=1
=>Phân số tối giản
\(\left(1\dfrac{3}{4}-\dfrac{4}{6}\right):\left(1\dfrac{1}{5}+2\dfrac{2}{5}+\dfrac{1}{5}\right)< x< 1\dfrac{1}{5}.1\dfrac{1}{4}+3\dfrac{2}{11}:2\dfrac{3}{121}\)
\(\Leftrightarrow\left(\dfrac{7}{4}-\dfrac{4}{6}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)< x< \dfrac{6}{5}.\dfrac{5}{4}+\dfrac{35}{11}:\dfrac{245}{121}\) \(\Leftrightarrow\left(\dfrac{21}{12}-\dfrac{8}{12}\right):\dfrac{19}{5}< x< \dfrac{3}{2}+\dfrac{35}{11}.\dfrac{121}{245}\) \(\Leftrightarrow\dfrac{13}{12}.\dfrac{5}{19}< x< \dfrac{3}{2}+\dfrac{2}{7}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{21}{14}+\dfrac{4}{14}\) \(\Leftrightarrow\dfrac{65}{228}< x< \dfrac{25}{14}\) \(\Leftrightarrow x=1\)
c) \(\dfrac{11}{10}-\dfrac{-7}{24}=\dfrac{11}{10}+\dfrac{7}{24}=\dfrac{167}{120}\)
e) \(\dfrac{-8}{3}\cdot\dfrac{15}{7}=\dfrac{-120}{21}=\dfrac{-40}{7}\)
f) \(\dfrac{-2}{5}\cdot4\dfrac{1}{2}=\dfrac{-2}{5}\cdot\dfrac{9}{2}=-\dfrac{9}{5}\)
g) \(\dfrac{5}{3}:\dfrac{5}{-3}=\dfrac{5}{3}:\dfrac{-5}{3}=\dfrac{5}{3}\cdot\dfrac{-3}{5}=-1\)
h) \(\dfrac{5}{4}:\left(-9\right)=\dfrac{5}{4}:\dfrac{-9}{1}=\dfrac{5}{4}\cdot\dfrac{-1}{9}=-\dfrac{5}{36}\)