3x^2>6 eching

$$3 x ^ { 2 } > 6$$

x uchun yechish

$$x\in \left(-\infty,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty\right)$$

Yechim qadamlari

Ikki tarafini 3 ga bo‘ling. 3 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$x^{2}>\frac{6}{3}$$

2 ni olish uchun 6 ni 3 ga bo‘ling.

$$x^{2}>2$$

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

$$x^{2}>\left(\sqrt{2}\right)^{2}$$

Tengsizlikda |x|>\sqrt{2} bor.

$$|x|>\sqrt{2}$$

|x|>\sqrt{2} ni x<-\sqrt{2}\text{; }x>\sqrt{2} sifatida qaytadan yozish.

$$x<-\sqrt{2}\text{; }x>\sqrt{2}$$