x^2+2x+6leq6+left(3+xright)^2 eching

$${ x }^{ 2 } +2x+6 \leq 6+ { \left(3+x \right) }^{ 2 }$$

x uchun yechish

$$x\geq -\frac{9}{4}$$

Yechim qadamlari

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(3+x\right)^{2} kengaytirilishi uchun ishlating.

$$x^{2}+2x+6\leq 6+9+6x+x^{2}$$

15 olish uchun 6 va 9'ni qo'shing.

$$x^{2}+2x+6\leq 15+6x+x^{2}$$

Ikkala tarafdan 6x ni ayirish.

$$x^{2}+2x+6-6x\leq 15+x^{2}$$

-4x ni olish uchun 2x va -6x ni birlashtirish.

$$x^{2}-4x+6\leq 15+x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

$$x^{2}-4x+6-x^{2}\leq 15$$

0 ni olish uchun x^{2} va -x^{2} ni birlashtirish.

$$-4x+6\leq 15$$

Ikkala tarafdan 6 ni ayirish.

$$-4x\leq 15-6$$

9 olish uchun 15 dan 6 ni ayirish.

$$-4x\leq 9$$

Ikki tarafini -4 ga bo‘ling. -4 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$x\geq -\frac{9}{4}$$