36-4*(a^2+1)5geq0 eching

$$36 - 4 \cdot ( a ^ { 2 } + 1 ) 5 \geq 0$$

a uchun yechish

$$a\in \begin{bmatrix}-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\end{bmatrix}$$

Yechim qadamlari

20 hosil qilish uchun 4 va 5 ni ko'paytirish.

$$36-20\left(a^{2}+1\right)\geq 0$$

-20 ga a^{2}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$36-20a^{2}-20\geq 0$$

16 olish uchun 36 dan 20 ni ayirish.

$$16-20a^{2}\geq 0$$

16-20a^{2} musbatida eng katta quvvatni koeffitsientini aniqlash uchun tengsizlikni -1 ga koʻpaytiring. -1 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$-16+20a^{2}\leq 0$$

\frac{4}{5} ni ikki tarafga qo’shing.

$$a^{2}\leq \frac{4}{5}$$

\frac{4}{5} ning kvadrat ildizini hisoblab, \frac{2\sqrt{5}}{5} natijaga ega bo‘ling. \frac{4}{5} ni \left(\frac{2\sqrt{5}}{5}\right)^{2} sifatida qaytadan yozish.

$$a^{2}\leq \left(\frac{2\sqrt{5}}{5}\right)^{2}$$

Tengsizlikda |a|\leq \frac{2\sqrt{5}}{5} bor.

$$|a|\leq \frac{2\sqrt{5}}{5}$$

|a|\leq \frac{2\sqrt{5}}{5} ni a\in \left[-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\right] sifatida qaytadan yozish.

$$a\in \begin{bmatrix}-\frac{2\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\end{bmatrix}$$