k^2-13geq0 eching

$$k ^ { 2 } - 13 \geq 0$$

k uchun yechish

$$k\in (-\infty,-\sqrt{13}]\cup [\sqrt{13},\infty)$$

Yechim qadamlari

13 ni ikki tarafga qo’shing.

$$k^{2}\geq 13$$

13 ning kvadrat ildizini hisoblab, \sqrt{13} natijaga ega bo‘ling. 13 ni \left(\sqrt{13}\right)^{2} sifatida qaytadan yozish.

$$k^{2}\geq \left(\sqrt{13}\right)^{2}$$

Tengsizlikda |k|\geq \sqrt{13} bor.

$$|k|\geq \sqrt{13}$$

|k|\geq \sqrt{13} ni k\leq -\sqrt{13}\text{; }k\geq \sqrt{13} sifatida qaytadan yozish.

$$k\leq -\sqrt{13}\text{; }k\geq \sqrt{13}$$