cho biểu thức P=( $\frac{x+1}{3x^2+3x}$ +$\frac{1-2x}{6x^2-3x}$ -1): $\frac{1-x}{2x}$ rút gọn biểu thức P tìm x thuộc X thuộc Z để P có giá trị nguyên Tìm x để P $\leq$ 1

Đáp án:

\(\begin{array}{l}
a) - \dfrac{{2x}}{{1 - x}}\\
b)\left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
c) - 1 \le x < 1;x \ne \left\{ {0;\dfrac{1}{2}} \right\}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left( { - 1;0;\dfrac{1}{2};1} \right)\\
P = \left( {\dfrac{{x + 1}}{{3{x^2} + 3x}} + \dfrac{{1 - 2x}}{{6{x^2} - 3x}} - 1} \right):\dfrac{{1 - x}}{{2x}}\\
 = \left( {\dfrac{{x + 1}}{{3x\left( {x + 1} \right)}} + \dfrac{{1 - 2x}}{{3x\left( {2x - 1} \right)}} - 1} \right):\dfrac{{1 - x}}{{2x}}\\
 = \left( {\dfrac{{x + 1}}{{3x\left( {x + 1} \right)}} - \dfrac{{2x - 1}}{{3x\left( {2x - 1} \right)}} - 1} \right):\dfrac{{1 - x}}{{2x}}\\
 = \left( {\dfrac{1}{{3x}} - \dfrac{1}{{3x}} - 1} \right).\dfrac{{2x}}{{1 - x}}\\
 =  - \dfrac{{2x}}{{1 - x}}\\
b)P =  - \dfrac{{2x}}{{1 - x}} = \dfrac{{2x}}{{x - 1}} = \dfrac{{2\left( {x - 1} \right) + 2}}{{x - 1}}\\
 = 2 + \dfrac{2}{{x - 1}}\\
P \in Z \to \dfrac{2}{{x - 1}} \in Z\\
 \to x - 1 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 =  - 2\\
x - 1 = 1\\
x - 1 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 3\\
x =  - 1\left( l \right)\\
x = 2\\
x = 0\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
c)P \le 1\\
 \to \dfrac{{2x}}{{x - 1}} \le 1\\
 \to \dfrac{{2x - x + 1}}{{x - 1}} \le 0\\
 \to \dfrac{{x + 1}}{{x - 1}} \le 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \ge 0\\
x - 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \le 0\\
x - 1 > 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge  - 1\\
x < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le  - 1\\
x > 1
\end{array} \right.\left( l \right)
\end{array} \right.\\
 \to  - 1 \le x < 1;x \ne \left\{ {0;\dfrac{1}{2}} \right\}
\end{array}\)