(2a-5)^2/2-(a-3)^2+1/2geqa^2 eching

$$\frac { ( 2 a - 5 ) ^ { 2 } } { 2 } - ( a - 3 ) ^ { 2 } + \frac { 1 } { 2 } \geq a ^ { 2 }$$

a uchun yechish

$$a\leq 1$$

Yechim qadamlari

Tenglamaning ikkala tarafini 2 ga ko'paytirish. 2 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$2\left(\frac{\left(2a-5\right)^{2}}{2}-\left(a-3\right)^{2}\right)+1\geq 2a^{2}$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(2a-5\right)^{2} kengaytirilishi uchun ishlating.

$$2\left(\frac{4a^{2}-20a+25}{2}-\left(a-3\right)^{2}\right)+1\geq 2a^{2}$$

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

$$2\left(\frac{4a^{2}-20a+25}{2}-\left(a^{2}-6a+9\right)\right)+1\geq 2a^{2}$$

a^{2}-6a+9 teskarisini topish uchun har birining teskarisini toping.

$$2\left(\frac{4a^{2}-20a+25}{2}-a^{2}+6a-9\right)+1\geq 2a^{2}$$

2 ga \frac{4a^{2}-20a+25}{2}-a^{2}+6a-9 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2\times \frac{4a^{2}-20a+25}{2}-2a^{2}+12a-18+1\geq 2a^{2}$$

2\times \frac{4a^{2}-20a+25}{2} ni yagona kasrga aylantiring.

$$\frac{2\left(4a^{2}-20a+25\right)}{2}-2a^{2}+12a-18+1\geq 2a^{2}$$

2 va 2 ni qisqartiring.

$$4a^{2}-20a+25-2a^{2}+12a-18+1\geq 2a^{2}$$

2a^{2} ni olish uchun 4a^{2} va -2a^{2} ni birlashtirish.

$$2a^{2}-20a+25+12a-18+1\geq 2a^{2}$$

-8a ni olish uchun -20a va 12a ni birlashtirish.

$$2a^{2}-8a+25-18+1\geq 2a^{2}$$

7 olish uchun 25 dan 18 ni ayirish.

$$2a^{2}-8a+7+1\geq 2a^{2}$$

8 olish uchun 7 va 1'ni qo'shing.

$$2a^{2}-8a+8\geq 2a^{2}$$

Ikkala tarafdan 2a^{2} ni ayirish.

$$2a^{2}-8a+8-2a^{2}\geq 0$$

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

$$-8a+8\geq 0$$

Ikkala tarafdan 8 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$-8a\geq -8$$

Ikki tarafini -8 ga bo‘ling. -8 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$a\leq \frac{-8}{-8}$$

1 ni olish uchun -8 ni -8 ga bo‘ling.

$$a\leq 1$$