Question

Tìm x, y nguyên để các biểu thức sau có giá trị nguyên:

a) F(x) = \(\dfrac{5x-2}{x-1}\)

b) G(x) = \(\dfrac{2x-7}{x+3}\)

c) B(n) = \(\dfrac{6n+5}{2n-1}\)

Answer 1

Giải:

a) Để biểu thức nguyên thì:

\(F\left(x\right)\in Z\)

\(\Leftrightarrow\dfrac{5x-2}{x-1}\in Z\)

\(\Leftrightarrow5x-2⋮x-1\)

\(\Leftrightarrow5x-5+3⋮x-1\)

\(\Leftrightarrow5\left(x-1\right)+3⋮x-1\)

\(\Leftrightarrow3⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow x=\left\{0;2;-2;4\right\}\)

Vậy ...

b) Để biểu thức nguyên thì:

\(G\left(x\right)\in Z\)

\(\Leftrightarrow\dfrac{2x-7}{x+3}\in Z\) \(\Leftrightarrow2x-7⋮x+3\) \(\Leftrightarrow2x+6-13⋮x+3\) \(\Leftrightarrow2\left(x+3\right)-13⋮x+3\) \(\Leftrightarrow-13⋮x+3\) \(\Leftrightarrow x+3\inƯ\left(-13\right)=\left\{\pm1;\pm13\right\}\) \(\Leftrightarrow x=\left\{-2;-4;-16;10\right\}\) c) Để biểu thức nguyên thì:

\(B\left(n\right)\in Z\)

\(\Leftrightarrow\dfrac{6n+5}{2n-1}\in Z\) \(\Leftrightarrow6n+5⋮2n-1\) \(\Leftrightarrow6n-3+8⋮2n-1\) \(\Leftrightarrow3\left(2n-1\right)+8⋮2n-1\) \(\Leftrightarrow8⋮2n-1\) \(\Leftrightarrow2n-1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\) \(\Leftrightarrow n\in\left\{0;1;-\dfrac{1}{2};\dfrac{3}{2};-\dfrac{3}{2};\dfrac{5}{2};\dfrac{9}{2};-\dfrac{7}{2}\right\}\) \(\Leftrightarrow n=\left\{0;1\right\}\) Vậy ...