Đáp án:
1) \(x \in \left[ {\dfrac{1}{4};\dfrac{1}{2}} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne \dfrac{1}{2}\\
\dfrac{{2{x^2} + x}}{{1 - 2x}} \ge 1 - x\\
\to \dfrac{{2{x^2} + x - \left( {1 - 2x} \right)\left( {1 - x} \right)}}{{1 - 2x}} \ge 0\\
\to \dfrac{{2{x^2} + x - 2{x^2} + 3x - 1}}{{1 - 2x}} \ge 0\\
\to \dfrac{{4x - 1}}{{1 - 2x}} \ge 0
\end{array}\)
BXD
x -∞ 1/4 1/2 +∞
f(x) - 0 + // -
\( \to x \in \left[ {\dfrac{1}{4};\dfrac{1}{2}} \right)\)
\(\begin{array}{l}
2)DK:x \ne 3\\
\dfrac{{4x - 7 + 2\left( {3 - x} \right)}}{{3 - x}} \ge 0\\
\to \dfrac{{4x - 7 + 6 - 2x}}{{3 - x}} \ge 0\\
\to \dfrac{{2x - 1}}{{3 - x}} \ge 0
\end{array}\)
BXD
x -∞ 1/2 3 +∞
f(x) - 0 + // -
\( \to x \in \left[ {\dfrac{1}{2};3} \right)\)
\(\begin{array}{l}
3)DK:x \ne \left\{ { - \dfrac{1}{2};2} \right\}\\
\dfrac{{\left( {x + 5} \right)\left( {2x + 1} \right) - \left( {x - 3} \right)\left( {2x - 4} \right)}}{{\left( {2x - 4} \right)\left( {2x + 1} \right)}} < 0\\
\to \dfrac{{2{x^2} + 11x + 5 - 2{x^2} + 14x - 12}}{{\left( {2x - 4} \right)\left( {2x + 1} \right)}} < 0\\
\to \dfrac{{25x - 7}}{{\left( {2x - 4} \right)\left( {2x + 1} \right)}} < 0
\end{array}\)
BXD
x -∞ -1/2 7/25 2 +∞
f(x) - // + 0 - // +
\( \to x \in \left( { - \infty ; - \dfrac{1}{2}} \right) \cup \left( {\dfrac{7}{{25}};2} \right)\)
