2-(1+x)(1+x)<x(2-x) eching

$$2 - ( 1 + x ) ( 1 + x ) < x ( 2 - x )$$

x uchun yechish

$$x>\frac{1}{4}$$

Yechim qadamlari

\left(1+x\right)^{2} hosil qilish uchun 1+x va 1+x ni ko'paytirish.

$$2-\left(1+x\right)^{2}<x\left(2-x\right)$$

\left(a+b\right)^{2}=a^{2}+2ab+b^{2} binom teoremasini \left(1+x\right)^{2} kengaytirilishi uchun ishlating.

$$2-\left(1+2x+x^{2}\right)<x\left(2-x\right)$$

1+2x+x^{2} teskarisini topish uchun har birining teskarisini toping.

$$2-1-2x-x^{2}<x\left(2-x\right)$$

1 olish uchun 2 dan 1 ni ayirish.

$$1-2x-x^{2}<x\left(2-x\right)$$

x ga 2-x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$1-2x-x^{2}<2x-x^{2}$$

Ikkala tarafdan 2x ni ayirish.

$$1-2x-x^{2}-2x<-x^{2}$$

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

$$1-4x-x^{2}<-x^{2}$$

x^{2} ni ikki tarafga qo’shing.

$$1-4x-x^{2}+x^{2}<0$$

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

$$1-4x<0$$

Ikkala tarafdan 1 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$-4x<-1$$

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

$$x>\frac{-1}{-4}$$

Ikkala surat va maxrajdan manfiy belgini olib tashlash bilan \frac{-1}{-4} kasrini \frac{1}{4} ga soddalashtirish mumkin.

$$x>\frac{1}{4}$$