$\frac{(2\left | x \right |-3)^{2}-\left | x \right |-6}{4x+1}\geq 0$
Edited by rovklee, 21-07-2012 - 17:13.
$\frac{(2\left | x \right |-3)^{2}-\left | x \right |-6}{4x+1}\geq 0$
Started By rovklee, 21-07-2012 - 17:12
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Posted 21-07-2012 - 17:12
rovklee
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Giải bất phương trình:
$\frac{(2\left | x \right |-3)^{2}-\left | x \right |-6}{4x+1}\geq 0$
$\frac{(2\left | x \right |-3)^{2}-\left | x \right |-6}{4x+1}\geq 0$
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