solucion de desigualdades 4X-1>16X+14,
4X-4<12
6X+5<2X-15,
9X+14<6x-12,
X²-2X-15>0,
X²+x-6>0,
X²-6X+5>0​

Respuestas

Respuesta dada por: kavitarajpoot1982
0

Respuesta:

3X²-12≥0

\begin{gathered}3 {x}^{2} = 12 \\ {x}^{2} = 4 \\ x = \sqrt{4 } \\ x = 2\end{gathered}

3x

2

=12

x

2

=4

x=

4

x=2

X² ≤ 4x+ 12

\begin{gathered} {x}^{2} \leqslant 4x + 12 \\ {x}^{2} - 4x - 12 \\ (x - 6)(x + 2) \leqslant 0 \\1) x - 6 \leqslant 0 \\ x \leqslant 6 \\ 2)x + 2 \leqslant 0 \\ x + \leqslant - 2\end{gathered}

x

2

⩽4x+12

x

2

−4x−12

(x−6)(x+2)⩽0

1)x−6⩽0

x⩽6

2)x+2⩽0

x+⩽−2

16X²-64>0

\begin{gathered}16 {x}^{2} - 64 > 0 \\ {x}^{2} = 4 \\ x = 2\end{gathered}

16x

2

−64>0

x

2

=4

x=2

6X²-5<x

\begin{gathered}6 {x}^{2} -x - 5 < 0 \\ (6x -5)(x + 1) < 0 \\ 1)6x - 5 < 0 \\ x = \frac{5}{6} \\ 2)x + 1 < 0 \\ x < -1\end{gathered}

6x

2

−x−5<0

(6x−5)(x+1)<0

1)6x−5<0

x=

6

5

2)x+1<0

x<−1

X²-2x-4≥3

\begin{gathered} {x}^{2} - 2x - 7 \geqslant 0 \\ \end{gathered}

x

2

−2x−7⩾0

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