(x-1)^2<3 eching

$$( x - 1 ) ^ { 2 } < 3$$

x uchun yechish

$$x\in \left(1-\sqrt{3},\sqrt{3}+1\right)$$

Yechim qadamlari

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.

$$\left(x-1\right)^{2}=0$$

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

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

Hisoblarni amalga oshiring.

$$x=\frac{2±2\sqrt{3}}{2}$$

x=\frac{2±2\sqrt{3}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\sqrt{3}+1 x=1-\sqrt{3}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\left(x-\left(\sqrt{3}+1\right)\right)\left(x-\left(1-\sqrt{3}\right)\right)<0$$

Koʻpaytma manfiy boʻlishi uchun x-\left(\sqrt{3}+1\right) va x-\left(1-\sqrt{3}\right) qarama-qarshi belgilar boʻlishi kerak. x-\left(\sqrt{3}+1\right) musbat, x-\left(1-\sqrt{3}\right) manfiy boʻlganda, yechimni toping.

$$x-\left(\sqrt{3}+1\right)>0 x-\left(1-\sqrt{3}\right)<0$$

Bu har qanday x uchun xato.

$$x\in \emptyset$$

x-\left(1-\sqrt{3}\right) musbat, x-\left(\sqrt{3}+1\right) manfiy boʻlganda, yechimni toping.

$$x-\left(1-\sqrt{3}\right)>0 x-\left(\sqrt{3}+1\right)<0$$

Ikkala tengsizlikning mos yechimi – x\in \left(1-\sqrt{3},\sqrt{3}+1\right).

$$x\in \left(1-\sqrt{3},\sqrt{3}+1\right)$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\in \left(1-\sqrt{3},\sqrt{3}+1\right)$$