\(\frac{15}{n}\)nhận giá trị nguyên <=>n thuộc Ư(15)
<=>n thuộc {1; -1; 3; -3; 5; -5; 15; -15}
Vậy \(\frac{15}{n}\)đạt giá trị nguyên <=>n thuộc {1; -1; 3; -3; 5; -5; 15; -15}
Để 3 phân số trên nhận giá trị nguyên thì
n\(\in\)Ư(15)=>n={\(\pm\)1;\(\pm\)3;\(\pm\)5;\(\pm\)15}
n+2\(\in\)Ư(12)
2n-5\(\in\)Ư(6)
=>n=\(\pm\)1;\(\pm\)3,...
\(\frac{12}{n+2}\)dật giá trị nguyên <=> 12 chia hết cho n+2
<=> n+2 thuộc Ư(12)
<=> n+2 thộc {-1; 1; -2; 2; -3; 3; -4; 4; -6; 6; -12; 12}
<=> n thuộc {-3; -1; -4; 0; -5; 1; -6; 2; -8; 4; -14; 10}
Vậy với n thuộc {-3; -1; -4; 0; -5; 1; -6; 2; -8; 4; -14; 10} thì \(\frac{12}{n+2}\)đạt giá trị nguyên
\(\frac{15}{n}\Rightarrow n\inƯ\left(15\right)=\left\{-1,-3,-5,-15,1,3,5,15\right\}\)
\(\frac{12}{n+2}\Rightarrow n+2\inƯ\left(12\right)=\left\{-1,-2,-6,-12,1,2,6,12\right\}\)
\(\Rightarrow n\in\left\{-3,-4,-8,-14,-1,0,4,10\right\}\)
\(\frac{6}{2n-5}\Rightarrow2n-5\inƯ\left(6\right)=\left\{-1,-2,-3,-6,1,2,3,6\right\}\)
Ta có bảng :
| 2n-5 | -1 | -2 | -3 | -6 | 1 | 2 | 4 | 6 |
| n | 2 | 3/2 (loại) | 1 | -1/2 (loại) | 3 | 7/2 (loại) | 9/2 (loại) | 11/5 (loại) |
Vậy \(n\in\left\{1,2,3\right\}\)
Bài giải
\(\frac{15}{n}\text{ nhận giá trị nguyên }\Rightarrow\text{ }15\text{ }⋮\text{ }n\text{ }\Rightarrow\text{ }n\inƯ\left(15\right)=\left\{\pm1\text{ ; }\pm3\text{ ; }\pm5\text{ ; }\pm15\right\}\)
\(\frac{12}{n+2}\text{ nhận giá trị nguyên }\Rightarrow\text{ }12\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }n\inƯ\left(12\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm4\text{ ; }\pm6\text{ ; }\pm12\right\}\)
\(\Rightarrow\text{ }n\in\left\{-3\text{ ; }-1\text{ ; }-4\text{ ; }0\text{ ; }-5\text{ ; }1\text{ ; }-6\text{ ; }2\text{ ; }-8\text{ ; }4\text{ ; }-14\text{ ; }10\right\}\)
\(\frac{6}{2n-5}\text{ nhận giá trị nguyên }\Rightarrow\text{ }6\text{ }⋮\text{ }2n-5\text{ }\Rightarrow\text{ }2n-5\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)
\(\Rightarrow\text{ }n\in\left\{2\text{ ; }3\text{ ; }\frac{3}{2}\text{(loại ) ; }\frac{7}{2}\text{( loại ) ; }1\text{ ; }4\text{ ; }-\frac{1}{2}\text{ loại ; }\frac{11}{2}\text{ ( loại ) }\right\}\)
\(\text{Vậy }n\in\left\{1\text{ ; }2\text{ ; }3\right\}\)
Bài giải
\(\frac{15}{n}\text{ nhận giá trị nguyên }\Rightarrow\text{ }15\text{ }⋮\text{ }n\text{ }\Rightarrow\text{ }n\inƯ\left(15\right)=\left\{\pm1\text{ ; }\pm3\text{ ; }\pm5\text{ ; }\pm15\right\}\)
\(\frac{12}{n+2}\text{ nhận giá trị nguyên }\Rightarrow\text{ }12\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }n\inƯ\left(12\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm4\text{ ; }\pm6\text{ ; }\pm12\right\}\)
\(\Rightarrow\text{ }n\in\left\{-3\text{ ; }-1\text{ ; }-4\text{ ; }0\text{ ; }-5\text{ ; }1\text{ ; }-6\text{ ; }2\text{ ; }-8\text{ ; }4\text{ ; }-14\text{ ; }10\right\}\)
\(\frac{6}{2n-5}\text{ nhận giá trị nguyên }\Rightarrow\text{ }6\text{ }⋮\text{ }2n-5\text{ }\Rightarrow\text{ }2n-5\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)
\(\Rightarrow\text{ }n\in\left\{2\text{ ; }3\text{ ; }\frac{3}{2}\text{(loại ) ; }\frac{7}{2}\text{( loại ) ; }1\text{ ; }4\text{ ; }-\frac{1}{2}\text{ loại ; }\frac{11}{2}\text{ ( loại ) }\right\}\)
\(\text{Vậy }n\in\left\{1\text{ ; }2\text{ ; }3\right\}\)
Bài giải
\(\frac{15}{n}\text{ nhận giá trị nguyên }\Rightarrow\text{ }15\text{ }⋮\text{ }n\text{ }\Rightarrow\text{ }n\inƯ\left(15\right)=\left\{\pm1\text{ ; }\pm3\text{ ; }\pm5\text{ ; }\pm15\right\}\)
\(\frac{12}{n+2}\text{ nhận giá trị nguyên }\Rightarrow\text{ }12\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }n\inƯ\left(12\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm4\text{ ; }\pm6\text{ ; }\pm12\right\}\)
\(\Rightarrow\text{ }n\in\left\{-3\text{ ; }-1\text{ ; }-4\text{ ; }0\text{ ; }-5\text{ ; }1\text{ ; }-6\text{ ; }2\text{ ; }-8\text{ ; }4\text{ ; }-14\text{ ; }10\right\}\)
\(\frac{6}{2n-5}\text{ nhận giá trị nguyên }\Rightarrow\text{ }6\text{ }⋮\text{ }2n-5\text{ }\Rightarrow\text{ }2n-5\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)
\(\Rightarrow\text{ }n\in\left\{2\text{ ; }3\text{ ; }\frac{3}{2}\text{(loại ) ; }\frac{7}{2}\text{( loại ) ; }1\text{ ; }4\text{ ; }-\frac{1}{2}\text{ loại ; }\frac{11}{2}\text{ ( loại ) }\right\}\)
\(\text{Vậy }n\in\left\{1\text{ ; }2\text{ ; }3\right\}\)
Bài giải
\(\frac{15}{n}\text{ nhận giá trị nguyên }\Rightarrow\text{ }15\text{ }⋮\text{ }n\text{ }\Rightarrow\text{ }n\inƯ\left(15\right)=\left\{\pm1\text{ ; }\pm3\text{ ; }\pm5\text{ ; }\pm15\right\}\)
\(\frac{12}{n+2}\text{ nhận giá trị nguyên }\Rightarrow\text{ }12\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }n\inƯ\left(12\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm4\text{ ; }\pm6\text{ ; }\pm12\right\}\)
\(\Rightarrow\text{ }n\in\left\{-3\text{ ; }-1\text{ ; }-4\text{ ; }0\text{ ; }-5\text{ ; }1\text{ ; }-6\text{ ; }2\text{ ; }-8\text{ ; }4\text{ ; }-14\text{ ; }10\right\}\)
\(\frac{6}{2n-5}\text{ nhận giá trị nguyên }\Rightarrow\text{ }6\text{ }⋮\text{ }2n-5\text{ }\Rightarrow\text{ }2n-5\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)
\(\Rightarrow\text{ }n\in\left\{2\text{ ; }3\text{ ; }\frac{3}{2}\text{(loại ) ; }\frac{7}{2}\text{( loại ) ; }1\text{ ; }4\text{ ; }-\frac{1}{2}\text{ loại ; }\frac{11}{2}\text{ ( loại ) }\right\}\)
\(\text{Vậy }n\in\left\{1\text{ ; }2\text{ ; }3\right\}\)
Bài giải
\(\frac{15}{n}\text{ nhận giá trị nguyên }\Rightarrow\text{ }15\text{ }⋮\text{ }n\text{ }\Rightarrow\text{ }n\inƯ\left(15\right)=\left\{\pm1\text{ ; }\pm3\text{ ; }\pm5\text{ ; }\pm15\right\}\)
\(\frac{12}{n+2}\text{ nhận giá trị nguyên }\Rightarrow\text{ }12\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }n\inƯ\left(12\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm4\text{ ; }\pm6\text{ ; }\pm12\right\}\)
\(\Rightarrow\text{ }n\in\left\{-3\text{ ; }-1\text{ ; }-4\text{ ; }0\text{ ; }-5\text{ ; }1\text{ ; }-6\text{ ; }2\text{ ; }-8\text{ ; }4\text{ ; }-14\text{ ; }10\right\}\)
\(\frac{6}{2n-5}\text{ nhận giá trị nguyên }\Rightarrow\text{ }6\text{ }⋮\text{ }2n-5\text{ }\Rightarrow\text{ }2n-5\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)
\(\Rightarrow\text{ }n\in\left\{2\text{ ; }3\text{ ; }\frac{3}{2}\text{(loại ) ; }\frac{7}{2}\text{( loại ) ; }1\text{ ; }4\text{ ; }-\frac{1}{2}\text{ loại ; }\frac{11}{2}\text{ ( loại ) }\right\}\)
\(\text{Vậy }n\in\left\{1\text{ ; }2\text{ ; }3\right\}\)
Bài giải
\(\frac{15}{n}\text{ nhận giá trị nguyên }\Rightarrow\text{ }15\text{ }⋮\text{ }n\text{ }\Rightarrow\text{ }n\inƯ\left(15\right)=\left\{\pm1\text{ ; }\pm3\text{ ; }\pm5\text{ ; }\pm15\right\}\)
\(\frac{12}{n+2}\text{ nhận giá trị nguyên }\Rightarrow\text{ }12\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }n\inƯ\left(12\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm4\text{ ; }\pm6\text{ ; }\pm12\right\}\)
\(\Rightarrow\text{ }n\in\left\{-3\text{ ; }-1\text{ ; }-4\text{ ; }0\text{ ; }-5\text{ ; }1\text{ ; }-6\text{ ; }2\text{ ; }-8\text{ ; }4\text{ ; }-14\text{ ; }10\right\}\)
\(\frac{6}{2n-5}\text{ nhận giá trị nguyên }\Rightarrow\text{ }6\text{ }⋮\text{ }2n-5\text{ }\Rightarrow\text{ }2n-5\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)
\(\Rightarrow\text{ }n\in\left\{2\text{ ; }3\text{ ; }\frac{3}{2}\text{(loại ) ; }\frac{7}{2}\text{( loại ) ; }1\text{ ; }4\text{ ; }-\frac{1}{2}\text{ loại ; }\frac{11}{2}\text{ ( loại ) }\right\}\)
\(\text{Vậy }n\in\left\{1\text{ ; }2\text{ ; }3\right\}\)
Bài giải
\(\frac{15}{n}\text{ nhận giá trị nguyên }\Rightarrow\text{ }15\text{ }⋮\text{ }n\text{ }\Rightarrow\text{ }n\inƯ\left(15\right)=\left\{\pm1\text{ ; }\pm3\text{ ; }\pm5\text{ ; }\pm15\right\}\)
\(\frac{12}{n+2}\text{ nhận giá trị nguyên }\Rightarrow\text{ }12\text{ }⋮\text{ }n+2\text{ }\Rightarrow\text{ }n\inƯ\left(12\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm4\text{ ; }\pm6\text{ ; }\pm12\right\}\)
\(\Rightarrow\text{ }n\in\left\{-3\text{ ; }-1\text{ ; }-4\text{ ; }0\text{ ; }-5\text{ ; }1\text{ ; }-6\text{ ; }2\text{ ; }-8\text{ ; }4\text{ ; }-14\text{ ; }10\right\}\)
\(\frac{6}{2n-5}\text{ nhận giá trị nguyên }\Rightarrow\text{ }6\text{ }⋮\text{ }2n-5\text{ }\Rightarrow\text{ }2n-5\inƯ\left(6\right)=\left\{\pm1\text{ ; }\pm2\text{ ; }\pm3\text{ ; }\pm6\right\}\)
\(\Rightarrow\text{ }n\in\left\{2\text{ ; }3\text{ ; }\frac{3}{2}\text{(loại ) ; }\frac{7}{2}\text{( loại ) ; }1\text{ ; }4\text{ ; }-\frac{1}{2}\text{ loại ; }\frac{11}{2}\text{ ( loại ) }\right\}\)
\(\text{Vậy }n\in\left\{1\text{ ; }2\text{ ; }3\right\}\)
