Cho p/s \(\frac{8n-3}{2n+1}\)( n thuộc Z)
Tìm n để A= \(\frac{13}{5}\)
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DKXD cua phan thuc \(n\ne-9\)
\(\frac{7n-1}{n+9}=\frac{7n+63-64}{n+9}=\frac{7\left(n+9\right)-64}{n+9}=\frac{7\left(n+9\right)}{n+9}-\frac{64}{n+9}\)\(=7-\frac{64}{n+9}\)
De phan thuc dat gia tri nguyen => \(\frac{64}{n+9}\)nguyen
<=> \(64⋮n+9\)<=> \(n+9\in U\left(64\right)\)
<=> \(n+9\in\left\{-64;-32;-16;-8;-4;-2;-1;1;2;4;8;16;32;64\right\}\)
=> \(n\in\left\{-73;-41;-25;-17;-13;-11;-10;-7;-5;-1;7;23;55\right\}\)
Vì A, B, C thuộc Z nên tử chia hết cho mẫu, đặt phép chia ra
\(A=\frac{2n+1}{n+5}\inℤ\Leftrightarrow2n+1⋮n+5\)
\(\Rightarrow2n+10-9⋮n+5\)
\(\Rightarrow2\left(n+5\right)-9⋮n+5\)
\(2\left(n+5\right)⋮n+5\)
\(\Rightarrow9⋮n+5\)
\(\Rightarrow n+5\inƯ\left(9\right)\)
\(n\inℤ\Rightarrow n+5\inℤ\)
\(\Rightarrow n+5\in\left\{-1;1;-3;3;-9;9\right\}\)
\(\Rightarrow n\in\left\{-6;-4;-8;-2;-14;4\right\}\)
gọi d\(\in\)uc(2n+1,n+5)
\(\Rightarrow1\left(2n+1\right)-2\left(n+5\right)⋮d\)
\(\Rightarrow2n+1-2n-10⋮d\)
\(\Rightarrow-9⋮d\Rightarrow d\in u\left(-9\right)=\left\{1;-1;9;-9\right\}\)
Lập bảng:
\(2n+1\) | \(1\) | \(-1\) | \(9\) | \(-9\) |
\(n\) | \(0\) | \(-1\) | \(4\) | \(-5\) |
\(n\inℤ\Rightarrow n\in\left\{0;-1;4;-5\right\}\)
a)\(A=\frac{2n-5}{n+3}=\frac{2n+6-11}{n+3}=\frac{2n+6}{n+3}-\frac{11}{n+3}=2-\frac{11}{n+3}\)
\(2\in Z\Rightarrow\)Để \(A=2-\frac{11}{n+3}\in Z\)thì \(\frac{11}{n+3}\in Z\Rightarrow n+3\inƯ\left(11\right)\)
\(Ư\left(11\right)=\left(\pm1;\pm11\right)\Rightarrow n+3=\left(\pm1;\pm11\right)\)
*\(n+3=1\Rightarrow n=-2\)
*\(n+3=-1\Rightarrow n=-4\)
*\(n+3=11\Rightarrow n=8\)
*\(n+3=-11\Rightarrow n=-14\)
\(\frac{8n-3}{2n+1}=\frac{13}{5}\)
\(\Rightarrow\left(8n-3\right)\cdot5=\left(2n+1\right)\cdot13\)
\(\Rightarrow40n-15=26n+13\)
\(\Rightarrow40n-26n=13+15\)
\(\Rightarrow14n=28\)
\(\Rightarrow n=28\div2\)
\(\Rightarrow n=14\)
ta có : 8n-3/2n+1=13/5
(8n-3).5=(2n+1).13
40n-15=26n+13
40n-26n=15+23
14n=28
suy ra n=28:14=2
vậy n=2