Bài 1(2,5 điểm) Cho P =(x+x^2)/(x^2-x+1):[(x+1)/x-1/(1-x)+(2-x^2)/(x^2-x)] a) Rút gọn P; b) Tìm x biết P =(3x-2)/(x-1) c) Tìm x để P < 0

Đáp án:

c) \(x < 1;x \ne 0\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ {0;1} \right\}\\
P = \dfrac{{{x^2} + x}}{{{x^2} – x + 1}}:\left[ {\dfrac{{x + 1}}{x} – \dfrac{1}{{1 – x}} + \dfrac{{2 – {x^2}}}{{{x^2} – x}}} \right]\\
 = \dfrac{{x\left( {x + 1} \right)}}{{{x^2} – x + 1}}:\left[ {\dfrac{{\left( {x + 1} \right)\left( {x – 1} \right) + x + 2 – {x^2}}}{{x\left( {x – 1} \right)}}} \right]\\
 = \dfrac{{x\left( {x + 1} \right)}}{{{x^2} – x + 1}}.\dfrac{{x\left( {x – 1} \right)}}{{{x^2} – 1 + x + 2 – {x^2}}}\\
 = \dfrac{{x\left( {x + 1} \right)}}{{{x^2} – x + 1}}.\dfrac{{x\left( {x – 1} \right)}}{{x + 1}}\\
 = \dfrac{{{x^2}\left( {x – 1} \right)}}{{{x^2} – x + 1}}\\
c)P < 0\\
 \to \dfrac{{{x^2}\left( {x – 1} \right)}}{{{x^2} – x + 1}} < 0\\
Do:\left\{ \begin{array}{l}
{x^2} > 0\forall x \ne \left\{ {0;1} \right\}\\
{x^2} – x + 1 > 0\forall x \ne \left\{ {0;1} \right\}
\end{array} \right.\\
 \to P < 0\\
 \Leftrightarrow x – 1 < 0\\
 \to x < 1;x \ne 0
\end{array}\)