5 bài tập Giải bất phương trình (có lời giải)
5 câu hỏi
a) \(\frac{{4x - 1}}{9} < \frac{{5 - 3x}}{6}\); b) \(\frac{{2x - 5}}{{18}} < \frac{{4x + 3}}{{10}}\).
a) \(\frac{{4x - 1}}{9} < \frac{{5 - 3x}}{6}\) \(\frac{{2\left( {4x - 1} \right)}}{{18}} < \frac{{3\left( {5 - 3x} \right)}}{{18}}\) \(8x - 2 < 15 - 9x\) \(\begin{array}{l}8x + 9x < 15 + 2\\17x < 17\\x < 1\end{array}\) Vậy \(x < 1\). | b)\(\frac{{2x - 5}}{{18}} < \frac{{4x + 3}}{{10}}\) \(\begin{array}{l}\frac{{5\left( {2x - 5} \right)}}{{90}} < \frac{{9\left( {4x + 3} \right)}}{{90}}\\10x - 25 < 36x + 27\end{array}\) \(\begin{array}{l}10x - 36x < 27 + 25\\ - 26x < 52\\x > \frac{{52}}{{ - 26}}\\x > - 2\end{array}\) Vậy \(x > - 2\). |
Giải các bất phương trình sau :
a) \(\frac{{5x + 2}}{5} < \frac{{4x - 3}}{4}\) b) \(\frac{{3\left( {2x + 1} \right)}}{{20}} + 1 < \frac{{3x + 13}}{{10}}\).
a) \(\frac{{5x + 2}}{5} < \frac{{4x - 3}}{4}\) \(\begin{array}{l}\frac{{4\left( {5x + 2} \right)}}{{20}} < \frac{{5\left( {4x - 3} \right)}}{{20}}\\20x + 8 < 20x - 15\\20x - 20x < - 8 - 15\\0x < - 23\end{array}\) Vậy bất phương trình này vô nghiệm. | b) \(\frac{{3\left( {2x + 1} \right)}}{{20}} + 1 < \frac{{3x + 13}}{{10}}\) \[\begin{array}{l}\frac{{3\left( {2x + 1} \right)}}{{20}} + \frac{{20}}{{20}} < \frac{{2\left( {3x + 13} \right)}}{{20}}\\6x + 3 + 20 < 6x + 26\\6x - 6x < 26 - 3 - 20\\0x < 3\end{array}\] Vậy bất phương trình này có nghiệm bất kì. |
Ta có \(\frac{{3x + 17}}{{10}} > \frac{{5x + 22}}{{15}}\) \(\begin{array}{l}\frac{{3\left( {3x + 17} \right)}}{{30}} > \frac{{2\left( {5x + 22} \right)}}{{30}}\\9x + 51 > 10x + 44\\9x - 10x > 44 - 51\end{array}\) \(\begin{array}{l} - x > - 7\\x < 7\left( {\rm{*}} \right)\end{array}\) | Ta có \(\frac{{x - 4}}{{30}} - 1 > \frac{{2x - 27}}{{24}}\) \(\begin{array}{l}\frac{{4\left( {x - 4} \right)}}{{120}} - \frac{{120}}{{120}} > \frac{{5\left( {2x - 27} \right)}}{{120}}\\4x - 16 - 120 > 10x - 135\end{array}\) \(\begin{array}{l}4x - 10x > 16 + 120 - 135\\ - 6x > 1\end{array}\) \(x < - \frac{1}{6}\left( {{\rm{**}}} \right)\) |
Ta có \(\frac{{2x + 4}}{3} - \frac{{4x - 7}}{{18}} > \frac{{2x - 5}}{9} - \frac{{2x - 1}}{{15}}\)
\(\begin{array}{l}\frac{{30\left( {2x + 4} \right)}}{{90}} - \frac{{5\left( {4x - 7} \right)}}{{90}} > \frac{{10\left( {2x - 5} \right)}}{{90}} - \frac{{6\left( {2x - 1} \right)}}{{90}}\\60x + 120 - 20x + 35 > 20x - 50 - 12x + 6\end{array}\)
\(\begin{array}{l}60x - 20x - 20x + 12x > - 120 - 35 - 50 + 6\\32x > - 199\\x > \frac{{ - 199}}{{32}}\\x > - 6\frac{7}{{32}}\end{array}\)
Vì \(x\) là số nguyên âm nên \[x \in \left\{ { - 6\,;\, - 5\,;\, - 4\,;\, - 3\,;\, - 2\,;\, - 1} \right\}\].
ĐKХĐ : \(x \ne - 3\)
Ta có \(\frac{{3x - 1}}{{x + 3}} > 2\)
\(\begin{array}{l}\frac{{3x - 1}}{{x + 3}} - 2 > 0\\\frac{{3x - 1 - 2x - 6}}{{x + 3}} > 0\\\frac{{x - 7}}{{x + 3}} > 0\end{array}\)
TH1: \(\left\{ \begin{array}{l}x - 7 > 0\\x + 3 > 0\end{array} \right.\) nên \(\left\{ \begin{array}{l}x > 7\\x > - 3\end{array} \right.\), do đó \(x > 7\).
TH2: \(\left\{ \begin{array}{l}x - 7 < 0\\x + 3 < 0\end{array} \right.\) nên \(\left\{ \begin{array}{l}x < 7\\x < - 3\end{array} \right.\) , do đó \(x > 7\) hoặc \(x < - 3\).
