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Để tích 2 PS là số nguyên thì 19⋮n-1 và n⋮9
⇒n-1∈Ư(19),9∈B(n)
⇒Ư(19)={\(\pm\)1;\(\pm\)19}
⇒n-1=1 ⇒n-1=19
⇒n-1=-1 ⇒n-1=-19
⇒n∈{2;20;0;-18} nhưng 9∈B(n)
⇒n∈{0;-18}
Giải:
Ta gọi tích hai số là A
Ta có:
\(A=\dfrac{19}{n-1}.\dfrac{n}{9}=\dfrac{19.n}{\left(n-1\right).9}\) (với n ≠ 1)
Vì \(ƯCLN\left(19;9\right)=1\) \(;ƯCLN\left(n;n-1\right)=1\)
\(\Rightarrow A\in Z\)
\(\Rightarrow n\in B\left(9\right)\) và \(\left(n-1\right)\inƯ\left(19\right)\)
Ta có bảng giá trị:
n-1 | 1 | -1 | 19 | -19 |
n | 2 | 0 | 20 | -18 |
\(\Rightarrow n\in\left\{-18;0\right\}\) (t/m)
Vậy \(n\in\left\{-18;0\right\}\)
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
\(\dfrac{7}{9}+\dfrac{1}{3} < x < \dfrac{43}{8}+\dfrac{1}{10}\)
\(\dfrac{10}{9} < x < \dfrac{219}{40}\)
Mà \(x \in N\)
\(=>x=\){`2;3;4;5`}
\(\dfrac{7}{9}+\dfrac{1}{3}< x< \dfrac{43}{8}+\dfrac{1}{10}\)
\(\dfrac{10}{9}< x< \dfrac{219}{40}\)
Mà \(x\inℕ\)
\(\Rightarrow\dfrac{10}{9}< 2\le x\le5< \dfrac{219}{40}\)
\(\Rightarrow2\le x\le5\)
\(\Rightarrow x\in\left\{2;3;4;5\right\}\)
Vậy: \(x\in\left\{2;3;4;5\right\}\)
\(a,\\ =>n-3\inƯ\left(-7\right)\\ Ư\left(-7\right)=\left\{1;-1;7;-7\right\}\\ =>\left\{{}\begin{matrix}n-3=1\\n-3=-1\\n-3=7\\n-3=-7\end{matrix}\right.\\ =>\left\{{}\begin{matrix}n=4\\n=2\\n=10\\n=-4\end{matrix}\right.\)
\(b,\dfrac{n-5}{n+1}=1-\dfrac{6}{n+1}\\ =>n+1\inƯ\left(6\right)\\ Ư\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\\ =>\left\{{}\begin{matrix}\left\{{}\begin{matrix}n+1=1\\n+1=-1\\n+1=2\\n+1=-2\end{matrix}\right.\\\left\{{}\begin{matrix}n+1=3\\n+1=-3\\n+1=6\\n+1=-6\end{matrix}\right.\end{matrix}\right.=>\left\{{}\begin{matrix}\left\{{}\begin{matrix}n=0\\n=-2\\n=1\\n=-3\end{matrix}\right.\\\left\{{}\begin{matrix}n=2\\n=-4\\n=5\\n=-7\end{matrix}\right.\end{matrix}\right.\)
\(A=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(A=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(A=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(A=\dfrac{28}{7}-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{4}{1}-\dfrac{31}{9}\)
\(A=\dfrac{36}{9}-\dfrac{31}{9}\)
\(\Rightarrow A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\)
\(B=\left(\dfrac{92}{9}+\dfrac{13}{5}\right)-\dfrac{56}{9}\)
\(B=\dfrac{92}{9}-\dfrac{56}{9}+\dfrac{13}{5}\)
\(B=\dfrac{36}{9}+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{20}{5}+\dfrac{13}{5}\)
\(\Rightarrow B=\dfrac{33}{5}\)
A = \(8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
A = \(8\dfrac{2}{7}-3\dfrac{4}{9}-4\dfrac{2}{7}\)\
A = \(\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
A = \(\left(8-4\right)-3\dfrac{4}{9}\)
A = \(4-3\dfrac{4}{9}\)
A = \(\dfrac{36}{9}-\dfrac{31}{9}\)
A = \(\dfrac{5}{9}\).
B = \(\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\)
B = \(10\dfrac{2}{9}+2\dfrac{3}{5}-6\dfrac{2}{9}\)
B = \(\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
B = \(4+2\dfrac{3}{5}\)
B = \(\dfrac{20}{5}+\dfrac{13}{5}\)
B = \(\dfrac{33}{5}\).
Bổ sung đề tìm n là số nguyên để 7/n+9 là số nguyên
Để 7/n+9 là số nguyên thì \(n+9\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{-8;-10;-2;-16\right\}\)