Question

ajutorrrrrr!!!!10 inecuatii​

Answer 1

Explicație pas cu pas:

1)

$4{x}^{2} + 2x - 2 > 0 \\ 2(2 {x}^{2} + x - 1) > 0 \\ 2(x + 1)(2x - 1) > 0 \\ 2(x + 1)(2x - 1) = 0$

$x_{1} = - 1 \\ x_{2} = - \frac{1}{2}$

$- \infty < x < - 1 \: sau \: \frac{1}{2} < x < + \infty$

2)

$- 2 {x}^{2} + 3x + 3 < 0$

$- 2 {x}^{2} + 3x + 3 = 0$

Δ = 3² + 4•2•3 = 9 + 24 = 33

$x_{1} = \frac{ - 3 + \sqrt{33} }{- 4} = \frac{3 - \sqrt{33} }{4} \\x_{2} = \frac{ - 3 - \sqrt{33} }{- 4} = \frac{3 + \sqrt{33} }{4}$

$- \infty < x < \frac{3 - \sqrt{33} }{4} \: sau \: \frac{3 + \sqrt{33} }{4} < x < + \infty$

3)

$- {x}^{2} + 4x - 9 \geqslant 0 \\ - (x - 2)^{2} - 5 \geqslant 0$

$fara \: solutii \: in \: multimea \: numerelor \: reale$

4)

$4{x}^{2} - 5x + 12 \leqslant 0$

$4{x}^{2} - 5x + 12 = 0$

Δ = (-5)² -4•4•12 = 25 - 192 = -167 < 0

fără soluții în mulțimea numerelor reale

5)

$- 2{x}^{2} - 3x + 1 \geqslant 0$

$- 2{x}^{2} - 3x + 1 = 0$

Δ = (-3)² + 4•2 = 9 + 8 = 17

$x_{1} = \frac{3 + \sqrt{17} }{ - 4} = - \frac{3 + \sqrt{17} }{4} \\ x_{1} = \frac{3 - \sqrt{17} }{ - 4} = - \frac{ 3 - \sqrt{17}}{4}$

$- \frac{3 - \sqrt{17}}{4} \leqslant x \leqslant - \frac{3 + \sqrt{17}}{4}$

6)

${x}^{2} + x + 1 > 0$

${\left(x + \frac{1}{2} \right)}^{2} + \frac{3}{4} > 0$

$- \infty < x < + \infty$

7)

$- {x}^{2} + x - 1 < 0 \\ - {\left(x - \frac{1}{2} \right)}^{2} - \frac{3}{4} < 0$

$- \infty < x < + \infty$

8)

${x}^{2} - x - 1 \leqslant 0$

${x}^{2} - x - 1 = 0$

Δ = 1 + 4 = 5

$x_{1} = \frac{1 - \sqrt{5} }{2} \\ x_{1} = \frac{1 + \sqrt{5} }{2}$

$\frac{1 - \sqrt{5} }{2} \leqslant x \leqslant \frac{1 + \sqrt{5} }{2}$

9)

${x}^{2} - x + 1 \geqslant 0$

${\left(x - \frac{1}{2} \right)}^{2} + \frac{3}{4} \geqslant 0$

$- \infty < x < + \infty$

10)

$-{x}^{2} - x - 1 \geqslant 0$

$- {\left(x + \frac{1}{2} \right)}^{2} - \frac{3}{4} \geqslant 0$

fără soluții în mulțimea numerelor reale