Question

5) Resolver em R as inequações:

A) (X^2 - 5x + 5) < 1
B) (X^2 - 5x) > Ou igual a 6​

Answer 1

Resposta:

$a) \: {x}^{1} < 4 \\ {x}^{2} < 1 \\ \\ \\ b) \: {x}^{1} \geqslant 6\\ {x }^{2} \geqslant - 1$

Explicação passo-a-passo:

$a) \: ({x}^{2} - 5x + 5) < 1 \\ {x}^{2} - 5x + 5 - 1 < 0 \\ {x}^{2} - 5x + 4 < 0 \\ a = 1 \\ b = - 5 \\ c = 4 \\ x < \frac{ - ( - 5) + - \sqrt{ {( - 5)}^{2} - 4 \times 1 \times 4 } }{2 \times 1} \\ x < \frac{5 + - \sqrt{25 - 16} }{2} \\ x < \frac{5 + - \sqrt{9} }{2} \\ x < \frac{5 + - 3}{2} \\ {x}^{1} < \frac{5 + 3}{2} < \frac{8}{2} < 4 \\ {x}^{2} < \frac{5 - 3}{2} < \frac{2}{2} < 1 \\ \\ \\ b) \: ( {x}^{2} - 5x) \geqslant 6 \\ {x}^{2} - 5x - 6 \geqslant 0 \\ a = 1 \\ b = - 5 \\ c = - 6 \\ x \geqslant \frac{ - ( - 5) + - \sqrt{ {( - 5)}^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1} \\ x \geqslant \frac{5 + - \sqrt{25 + 24} }{2} \\ x \geqslant \frac{5 + - \sqrt{49} }{2} \\ x \geqslant \frac{5 + - 7}{2} \\ {x}^{1} \geqslant \frac{5 + 7}{2} \geqslant \frac{12}{2} \geqslant 6 \\ {x}^{2} \geqslant \frac{5 - 7}{2} \geqslant - \frac{2}{2} \geqslant - 1$