Question

Exercițiul 86 va rog mult

Answer 1

Sper sa te ajute. Succes

Answer 2

$x^{2} \leq 7-4 \sqrt{15-4 \sqrt{11} }$

$\sqrt{15-4 \sqrt{11} }= \sqrt{( \sqrt{11}-2)^{2} } = | \sqrt{11}-2 | = \sqrt{11}-2$

$x^{2} \leq 7-4*( \sqrt{11}-2) \ \textless \ =\ \textgreater \ x^{2} \leq 7-4 \sqrt{11}+8 \ \textless \ =\ \textgreater \ x^{2} \leq 15-4 \sqrt{11}$

$x^{2} \leq 15-4 \sqrt{11} \ \textless \ =\ \textgreater \ x^{2} \leq ( \sqrt{11}-2)^{2} \ \textless \ =\ \textgreater \ x \leq \sqrt{11}-2$

√11 ≈ 3,3 → o sa-l luam 3 ,deoarece x ∈ Z

[tex]x \leq 3-2 \ \textless \ =\ \textgreater \ x \leq 1 [/tex]

    x ∈ Z
⇒ x ∈ (-∞ ;1]