Để C có giá trị nguyên thì \(\frac{x+1}{2x-3}\) có giá trị nguyên
\(\Rightarrow x+1⋮2x-3\)
\(\Rightarrow2x+2⋮2x-3\)
\(\Rightarrow2x-3+5⋮2x-3\)
\(\Rightarrow5⋮2x-3\)
\(\Rightarrow2x-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow2x\in\left\{4;2;8;-2\right\}\)
\(\Rightarrow x\in\left\{2;1;4;-1\right\}\)
Vậy \(x\in\left\{-1;1;2;4\right\}\).
\(C=\frac{x+1}{2x-3}=\frac{2x+2}{2x-3}=\frac{2x-3+5}{2x-3}=\frac{5}{2x-3}\)
\(\Leftrightarrow2x-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
2x - 3 | 1 | -1 | 5 | -5 |
2x | 4 | 2 | 8 | -2 |
x | 2 | -1 | 4 | -1 |