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x^{2}\leq \frac{1}{5}
Ikki tarafini 5 ga bo‘ling. 5 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.
x^{2}\leq \left(\frac{\sqrt{5}}{5}\right)^{2}
\frac{1}{5} ning kvadrat ildizini hisoblab, \frac{\sqrt{5}}{5} natijaga ega bo‘ling. \frac{1}{5} ni \left(\frac{\sqrt{5}}{5}\right)^{2} sifatida qaytadan yozish.
|x|\leq \frac{\sqrt{5}}{5}
Tengsizlikda |x|\leq \frac{\sqrt{5}}{5} bor.
x\in \begin{bmatrix}-\frac{\sqrt{5}}{5},\frac{\sqrt{5}}{5}\end{bmatrix}
|x|\leq \frac{\sqrt{5}}{5} ni x\in \left[-\frac{\sqrt{5}}{5},\frac{\sqrt{5}}{5}\right] sifatida qaytadan yozish.