(x+1)(x-2)>(x-5)(x+5) eching

$$( x + 1 ) ( x - 2 ) > ( x - 5 ) ( x + 5 )$$

x uchun yechish

$$x<23$$

Yechim qadamlari

x+1 ga x-2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$x^{2}-x-2>\left(x-5\right)\left(x+5\right)$$

Hisoblang: \left(x-5\right)\left(x+5\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 5 kvadratini chiqarish.

$$x^{2}-x-2>x^{2}-25$$

Ikkala tarafdan x^{2} ni ayirish.

$$x^{2}-x-2-x^{2}>-25$$

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

$$-x-2>-25$$

2 ni ikki tarafga qo’shing.

$$-x>-25+2$$

-23 olish uchun -25 va 2'ni qo'shing.

$$-x>-23$$

Ikki tarafini -1 ga bo‘ling. -1 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$x<\frac{-23}{-1}$$

Ikkala surat va maxrajdan manfiy belgini olib tashlash bilan \frac{-23}{-1} kasrini 23 ga soddalashtirish mumkin.

$$x<23$$