(9-5x)^2+2(9-5x)^2-24<0 eching

$$( 9 - 5 x ) ^ { 2 } + 2 ( 9 - 5 x ) ^ { 2 } - 24 < 0$$

x uchun yechish

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(9-5x\right)^{2} kengaytirilishi uchun ishlating.

$$81-90x+25x^{2}+2\left(9-5x\right)^{2}-24<0$$

\left(a-b\right)^{2}=a^{2}-2ab+b^{2} binom teoremasini \left(9-5x\right)^{2} kengaytirilishi uchun ishlating.

$$81-90x+25x^{2}+2\left(81-90x+25x^{2}\right)-24<0$$

2 ga 81-90x+25x^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$81-90x+25x^{2}+162-180x+50x^{2}-24<0$$

243 olish uchun 81 va 162'ni qo'shing.

$$243-90x+25x^{2}-180x+50x^{2}-24<0$$

-270x ni olish uchun -90x va -180x ni birlashtirish.

$$243-270x+25x^{2}+50x^{2}-24<0$$

75x^{2} ni olish uchun 25x^{2} va 50x^{2} ni birlashtirish.

$$243-270x+75x^{2}-24<0$$

219 olish uchun 243 dan 24 ni ayirish.

$$219-270x+75x^{2}<0$$

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.

$$219-270x+75x^{2}=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 75 ni, b uchun -270 ni va c uchun 219 ni ayiring.

$$x=\frac{-\left(-270\right)±\sqrt{\left(-270\right)^{2}-4\times 75\times 219}}{2\times 75}$$

Hisoblarni amalga oshiring.

$$x=\frac{270±60\sqrt{2}}{150}$$

x=\frac{270±60\sqrt{2}}{150} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{2\sqrt{2}+9}{5} x=\frac{9-2\sqrt{2}}{5}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$75\left(x-\frac{2\sqrt{2}+9}{5}\right)\left(x-\frac{9-2\sqrt{2}}{5}\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\frac{2\sqrt{2}+9}{5} va x-\frac{9-2\sqrt{2}}{5} qarama-qarshi belgilar boʻlishi kerak. x-\frac{2\sqrt{2}+9}{5} musbat, x-\frac{9-2\sqrt{2}}{5} manfiy boʻlganda, yechimni toping.

$$x-\frac{2\sqrt{2}+9}{5}>0 x-\frac{9-2\sqrt{2}}{5}<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\frac{9-2\sqrt{2}}{5} musbat, x-\frac{2\sqrt{2}+9}{5} manfiy boʻlganda, yechimni toping.

$$x-\frac{9-2\sqrt{2}}{5}>0 x-\frac{2\sqrt{2}+9}{5}<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right).

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(\frac{9-2\sqrt{2}}{5},\frac{2\sqrt{2}+9}{5}\right)$$