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ineq et ensemble du 13 oct

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Debucquoy Anthony
2022-10-14 12:56:02 +02:00
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src/math/ineq/abs.md
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@@ -9,19 +9,19 @@
1) \\(\lvert 3x+5\rvert \leq 2\\)
- Si \\( 3x + 5 \geq 0 \iff x \geq \frac{-5}{3}\\)
- Alors \\(\lvert 3x+5\rvert \leq 2\\) devient \\(3x + 5 \leq 2\\)
- \\(\iff x \leq -1\\)
- Si \\( 3x + 5 \geq 0 \iff x \geq \frac{-5}{3}\\)
- Alors \\(\lvert 3x+5\rvert \leq 2\\) devient \\(3x + 5 \leq 2\\)
- \\(\iff x \leq -1\\)
- > Ensemble des solutions trouvées pour ce cas :
\\[ ]-\infty, -1] \cap \left[\frac{-5}{3}, +\infty\right[ = \left[\frac{-5}{3}, -1\right]\\]
- > Ensemble des solutions trouvées pour ce cas :
\\[ ]-\infty, -1] \cap \left[\frac{-5}{3}, +\infty\right[ = \left[\frac{-5}{3}, -1\right]\\]
- Si \\( 3x + 5 < 0 \iff x \leq \frac{-5}{3}\\)
- Alors \\(\lvert 3x+5\rvert \leq 2\\) devient \\-(3x + 5 \leq 2\\)
- \\(\iff x \leq \frac{-7}{3}\\)
- > Ensemble des solutions trouvées pour ce cas :
\\[ \left[\frac{-7}{3}, +\infty\right[\cap\left]-\infty, \frac{-5}{3}\right[ = \left[\frac{-7}{3}, \frac{-5}{3}\right[\\]
- Si \\( 3x + 5 < 0 \iff x \leq \frac{-5}{3}\\)
- Alors \\(\lvert 3x+5\rvert \leq 2\\) devient \\-(3x + 5 \leq 2\\)
- \\(\iff x \leq \frac{-7}{3}\\)
- > Ensemble des solutions trouvées pour ce cas :
\\[ \left[\frac{-7}{3}, +\infty\right[\cap\left]-\infty, \frac{-5}{3}\right[ = \left[\frac{-7}{3}, \frac{-5}{3}\right[\\]
- > Conclusion:
\\[\\{x \vert\lvert 3x+5\rvert \leq 2 \\} = \left[\frac{-5}{3}, -1\right] \cup \left[\frac{-7}{3}, \frac{-5}{3}\right[ = \left[\frac{-7}{3}, -1\right] \\]
- > Conclusion:
\\[\\{x \vert\lvert 3x+5\rvert \leq 2 \\} = \left[\frac{-5}{3}, -1\right] \cup \left[\frac{-7}{3}, \frac{-5}{3}\right[ = \left[\frac{-7}{3}, -1\right] \\]