Question

với những giá trị nguyên nào của X thì phân số sau tối giản

\(\dfrac{x-8}{2x-17}\)

\(\dfrac{x-4}{x+1}\)

\(\dfrac{10}{x+7}\)

\(\dfrac{x-1}{x^2}\)

Answer 1

\(x\in Z\Rightarrow2x-17\ne0;Goi:d=\left(x-8,2x-17\right)\Rightarrow\left\{{}\begin{matrix}x-8⋮d\\2x-17⋮d\end{matrix}\right.\Leftrightarrow2\left(x-8\right)-\left(2x-17\right)⋮d\Leftrightarrow1⋮d\Leftrightarrow d=1\Rightarrow\frac{x-8}{2x-17}toigian\forall x\)

\(ĐK:x\ne-1;Goi:d=\left(x-4,x+1\right)\Rightarrow\left\{{}\begin{matrix}x-4⋮d\\x+1⋮d\end{matrix}\right.\Rightarrow x+1-\left(x-4\right)⋮d\Leftrightarrow5⋮d\Rightarrow d\in\left\{1;5\right\}\) gia sư: \(x+1⋮5\Rightarrow x\ne5k+1\left(k\in N\right)\)