Qual o resultado da inequação (5x+2)(2-x)(4x+3)≥0 para eu poder conferir se acertei?
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Boa noite!
![(5x + 2)(2 - x)(4x + 3) \geqslant 0 \\ \\ (5x + 2)(2 - x) \geqslant 0 \\ \\ 4x + 3 \geqslant 0 \\ \\ (5x + 2)(2 - x) \leqslant 0 \\ \\ 4x + 3 \geqslant 0 \\ \\ (5x + 2)(2 - x) \leqslant 0 \\ \\ 4x + 3 \leqslant 0 \\ \\ xe( - \frac{2}{5} \: \: \: \: \: \: \: \: 2) \\ \\ x \geqslant - \frac{3}{4} \\ \\ xe( - \infty \: \: \: \: \: - \frac{3}{4} ) \\ \\ xe( - \infty \: \: \: \: \: - \frac{3}{4} )u( - \frac{2}{5} \: \: \: \: \: 2) (5x + 2)(2 - x)(4x + 3) \geqslant 0 \\ \\ (5x + 2)(2 - x) \geqslant 0 \\ \\ 4x + 3 \geqslant 0 \\ \\ (5x + 2)(2 - x) \leqslant 0 \\ \\ 4x + 3 \geqslant 0 \\ \\ (5x + 2)(2 - x) \leqslant 0 \\ \\ 4x + 3 \leqslant 0 \\ \\ xe( - \frac{2}{5} \: \: \: \: \: \: \: \: 2) \\ \\ x \geqslant - \frac{3}{4} \\ \\ xe( - \infty \: \: \: \: \: - \frac{3}{4} ) \\ \\ xe( - \infty \: \: \: \: \: - \frac{3}{4} )u( - \frac{2}{5} \: \: \: \: \: 2)](https://tex.z-dn.net/?f=%285x+%2B+2%29%282+-+x%29%284x+%2B+3%29+%5Cgeqslant+0+%5C%5C++%5C%5C+%285x+%2B+2%29%282+-+x%29+%5Cgeqslant+0+%5C%5C++%5C%5C+4x+%2B+3+%5Cgeqslant+0+%5C%5C++%5C%5C+%285x+%2B+2%29%282+-+x%29+%5Cleqslant+0+%5C%5C++%5C%5C+4x+%2B+3+%5Cgeqslant+0+%5C%5C++%5C%5C+%285x+%2B+2%29%282+-+x%29+%5Cleqslant+0+%5C%5C++%5C%5C+4x+%2B+3+%5Cleqslant+0+%5C%5C++%5C%5C+xe%28+-++%5Cfrac%7B2%7D%7B5%7D+%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+2%29+%5C%5C++%5C%5C+x+%5Cgeqslant++-++%5Cfrac%7B3%7D%7B4%7D++%5C%5C++%5C%5C+xe%28+-++%5Cinfty++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++-++%5Cfrac%7B3%7D%7B4%7D+%29+%5C%5C++%5C%5C++xe%28+-++%5Cinfty++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A++-++%5Cfrac%7B3%7D%7B4%7D+%29u%28+-++%5Cfrac%7B2%7D%7B5%7D++%5C%3A++%5C%3A++%5C%3A++%5C%3A++%5C%3A+2%29)
Boa noite!
mariocezar:
mais tarde eu ponho uma explicação ok
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