(50-(x-100/5))x-1100>0 eching

$$(50-( \frac{ x-100 }{ 5 } ))x-1100 > 0$$

x uchun yechish

$$x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Tenglamaning ikkala tarafini 5 ga ko'paytirish. 5 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$5\left(50-\frac{x-100}{5}\right)x-5500>0$$

5 ga 50-\frac{x-100}{5} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\left(250+5\left(-\frac{x-100}{5}\right)\right)x-5500>0$$

5\left(-\frac{x-100}{5}\right) ni yagona kasrga aylantiring.

$$\left(250+\frac{-5\left(x-100\right)}{5}\right)x-5500>0$$

5 va 5 ni qisqartiring.

$$\left(250-\left(x-100\right)\right)x-5500>0$$

x-100 teskarisini topish uchun har birining teskarisini toping.

$$\left(250-x-\left(-100\right)\right)x-5500>0$$

-100 ning teskarisi 100 ga teng.

$$\left(250-x+100\right)x-5500>0$$

350 olish uchun 250 va 100'ni qo'shing.

$$\left(350-x\right)x-5500>0$$

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

$$350x-x^{2}-5500>0$$

350x-x^{2}-5500 musbatida eng katta quvvatni koeffitsientini aniqlash uchun tengsizlikni -1 ga koʻpaytiring. -1 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$-350x+x^{2}+5500<0$$

Tengsizlikni yechish uchun chap tomon faktorini hisoblang. Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$-350x+x^{2}+5500=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 1 ni, b uchun -350 ni va c uchun 5500 ni ayiring.

$$x=\frac{-\left(-350\right)±\sqrt{\left(-350\right)^{2}-4\times 1\times 5500}}{2}$$

Hisoblarni amalga oshiring.

$$x=\frac{350±10\sqrt{1005}}{2}$$

x=\frac{350±10\sqrt{1005}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=5\sqrt{1005}+175 x=175-5\sqrt{1005}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(5\sqrt{1005}+175\right)\right)\left(x-\left(175-5\sqrt{1005}\right)\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\left(5\sqrt{1005}+175\right) va x-\left(175-5\sqrt{1005}\right) qarama-qarshi belgilar boʻlishi kerak. x-\left(5\sqrt{1005}+175\right) musbat, x-\left(175-5\sqrt{1005}\right) manfiy boʻlganda, yechimni toping.

$$x-\left(5\sqrt{1005}+175\right)>0 x-\left(175-5\sqrt{1005}\right)<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\left(175-5\sqrt{1005}\right) musbat, x-\left(5\sqrt{1005}+175\right) manfiy boʻlganda, yechimni toping.

$$x-\left(175-5\sqrt{1005}\right)>0 x-\left(5\sqrt{1005}+175\right)<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right).

$$x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(175-5\sqrt{1005},5\sqrt{1005}+175\right)$$