5x(x-1)-3>2x-x eching

$$5 x ( x - 1 ) - 3 > 2 x - x$$

x uchun yechish

$$x\in \left(-\infty,\frac{3-2\sqrt{6}}{5}\right)\cup \left(\frac{2\sqrt{6}+3}{5},\infty\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

5x ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x^{2}-5x-3>2x-x$$

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

$$5x^{2}-5x-3>x$$

Ikkala tarafdan x ni ayirish.

$$5x^{2}-5x-3-x>0$$

-6x ni olish uchun -5x va -x ni birlashtirish.

$$5x^{2}-6x-3>0$$

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.

$$5x^{2}-6x-3=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 5 ni, b uchun -6 ni va c uchun -3 ni ayiring.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Hisoblarni amalga oshiring.

$$x=\frac{6±4\sqrt{6}}{10}$$

x=\frac{6±4\sqrt{6}}{10} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{2\sqrt{6}+3}{5} x=\frac{3-2\sqrt{6}}{5}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$5\left(x-\frac{2\sqrt{6}+3}{5}\right)\left(x-\frac{3-2\sqrt{6}}{5}\right)>0$$

Koʻpaytma musbat boʻlishi uchun x-\frac{2\sqrt{6}+3}{5} va x-\frac{3-2\sqrt{6}}{5} ikkalasi yo manfiy, yo musbat boʻlishi kerak. x-\frac{2\sqrt{6}+3}{5} va x-\frac{3-2\sqrt{6}}{5} ikkalasi manfiy boʻlganda, yechimini toping.

$$x-\frac{2\sqrt{6}+3}{5}<0 x-\frac{3-2\sqrt{6}}{5}<0$$

Ikkala tengsizlikning mos yechimi – x<\frac{3-2\sqrt{6}}{5}.

$$x<\frac{3-2\sqrt{6}}{5}$$

x-\frac{2\sqrt{6}+3}{5} va x-\frac{3-2\sqrt{6}}{5} ikkalasi musbat boʻlganda, yechimini toping.

$$x-\frac{3-2\sqrt{6}}{5}>0 x-\frac{2\sqrt{6}+3}{5}>0$$

Ikkala tengsizlikning mos yechimi – x>\frac{2\sqrt{6}+3}{5}.

$$x>\frac{2\sqrt{6}+3}{5}$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<\frac{3-2\sqrt{6}}{5}\text{; }x>\frac{2\sqrt{6}+3}{5}$$