Giải các phương trình sau
a) \(\frac{3x+1}{2}-\frac{x-2}{3}< \frac{1-2x}{4};\)
b) \((2x - 1)(x + 3) - 3x + 1 \leq (x - 1)(x + 3) + x^2 - 5.\)
Câu a:
\(\frac{{3x + 1}}{2} - \frac{{x - 2}}{3} < \frac{{1 - 2x}}{4} \Leftrightarrow \frac{{3(3x + 1) - 2(x - 2)}}{6} - \frac{{1 - 2x}}{4} < 0\)
\(\Leftrightarrow \frac{{7x + 7}}{6} + \frac{{2x - 1}}{4} < 0 \Leftrightarrow 14x + 14 + 6x - 3 < 0 \Leftrightarrow 20x < - 11\)
\( \Leftrightarrow x < - \frac{{11}}{{20}}\)
Vậy \(S = \left( { - \infty ; - \frac{{11}}{{20}}} \right)\)
Câu b:
\((2x - 1)(x + 3) - 3x + 1 \le (x - 1)(x + 3) + {x^2} - 5\)
\(\Leftrightarrow 2{x^2} + 5x - 3 - 3x + 1 \le {x^2} + 2x - 3 + {x^2} - 5\)
\( \Leftrightarrow 1 \le - 5\) vô nghiệm.
\(S = \emptyset \)
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