Question

Va rog!!!!!!!!!!!!!!​

Answer 1

Explicație pas cu pas:

1)

$\frac{3x - 7}{5 - 7x} \leqslant 3 \\ \frac{3x - 7}{5 - 7x} - 3 \leqslant 0 \\ \frac{3x - 7 - 3(5 - 7x)}{5 - 7x} \leqslant 0 \\ \frac{3x - 7 - 15 + 21x}{5 - 7x} \leqslant 0 \\ \frac{24x - 22}{5 - 7x} \leqslant 0$

$24x - 22 = 0 = > x = \frac{11}{12} = >$

$24x - 22 \leqslant 0 = > x \leqslant \frac{11}{12} \\ 24x - 22 \geqslant0 = > x \geqslant \frac{11}{12}$

și

$5 - 7x = 0 = > x = \frac{5}{7}$

$5 - 7x < 0 = > x > \frac{5}{7} \\5 - 7x > 0 = > x < \frac{5}{7}$

=>

$- \infty < x < \frac{5}{7} \: sau \: \frac{11}{12} \leqslant x < + \infty$

2)

$\frac{ {x}^{2} - x + 6}{ {x}^{2} + x + 8} > 1 \\ \frac{ {x}^{2} - x + 6}{ {x}^{2} + x + 8} - 1> 0 \\ \frac{ {x}^{2} - x + 6 - ({x}^{2} + x + 8)}{ {x}^{2} + x + 8}> 0 \\ \frac{ {x}^{2} - x + 6 - {x}^{2} - x - 8}{ {x}^{2} + x + 8}> 0 \\ \frac{ - 2x - 2}{{x}^{2} + x + 8} > 0 \\ \frac{ - 2(x + 1)}{{x}^{2} + x + 8} > 0 \\ \frac{ 2(x + 1)}{{x}^{2} + x + 8} < 0$

x² + x + 8 > 0, x ∈ R

$x + 1 = 0 = > x = - 1$

$x + 1 < 0 < = > x < - 1 \\ x + 1 > 0 < = > x > - 1$

=>

$- \infty < x < - 1$