Cho biểu thức Q=(2x-x^2/2x^2+8 -2x^2/x^3-2x^2+4x-8) .(2/x^2 -x-1/x) a, Tìm đkxđ và rút gọn Q b, Tìm đk của x để Q>1/2 c, Tìm các giá trị nguyên

Đáp án:

 a) \(\dfrac{{x + 1}}{{2x}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ {0;2} \right\}\\
Q = \left( {\dfrac{{2x – {x^2}}}{{2{x^2} + 8}} – \dfrac{{2{x^2}}}{{{x^3} – 2{x^2} + 4x – 8}}} \right).\left( {\dfrac{2}{{{x^2}}} – \dfrac{{x – 1}}{x}} \right)\\
 = \left[ {\dfrac{{2x – {x^2}}}{{2{x^2} + 8}} – \dfrac{{2{x^2}}}{{{x^2}\left( {x – 2} \right) + 4\left( {x – 2} \right)}}} \right].\dfrac{{2 – {x^2} + x}}{{{x^2}}}\\
 = \left[ {\dfrac{{2x – {x^2}}}{{2\left( {{x^2} + 4} \right)}} – \dfrac{{2{x^2}}}{{\left( {{x^2} + 4} \right)\left( {x – 2} \right)}}} \right].\dfrac{{\left( {2 – x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \left[ {\dfrac{{\left( {2x – {x^2}} \right)\left( {x – 2} \right) – 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x – 2} \right)}}} \right].\dfrac{{\left( {2 – x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{ – {x^3} + 2{x^2} – 4x + 2{x^2} – 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x – 2} \right)}}.\dfrac{{\left( {2 – x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{ – {x^3} – 4x}}{{2\left( {{x^2} + 4} \right)\left( {x – 2} \right)}}.\dfrac{{\left( {2 – x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{ – x\left( {{x^2} + 4} \right)}}{{2\left( {{x^2} + 4} \right)\left( {x – 2} \right)}}.\dfrac{{\left( {2 – x} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
 = \dfrac{{x + 1}}{{2x}}\\
b)Q > \dfrac{1}{2}\\
 \to \dfrac{{x + 1}}{{2x}} > \dfrac{1}{2}\\
 \to \dfrac{{2x + 2 – 2x}}{{4x}} > 0\\
 \to \dfrac{2}{{4x}} > 0\\
 \to x > 0;x \ne 2\\
c)Q = \dfrac{{x + 1}}{{2x}}\\
 \to 2Q = \dfrac{{x + 1}}{x} = 1 + \dfrac{1}{x}\\
Q \in Z \Leftrightarrow \dfrac{1}{x} \in Z\\
 \Leftrightarrow x \in U\left( 1 \right)\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  – 1
\end{array} \right.
\end{array}\)