6x^2-12x+3geq0 eching

$$6 x ^ { 2 } - 12 x + 3 \geq 0$$

x uchun yechish

$$x\in (-\infty,-\frac{\sqrt{2}}{2}+1]\cup [\frac{\sqrt{2}}{2}+1,\infty)$$

Kvadrat tenglamalaridan foydalanish qoidalari

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.

$$6x^{2}-12x+3=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 6 ni, b uchun -12 ni va c uchun 3 ni ayiring.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 6\times 3}}{2\times 6}$$

Hisoblarni amalga oshiring.

$$x=\frac{12±6\sqrt{2}}{12}$$

x=\frac{12±6\sqrt{2}}{12} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$x=\frac{\sqrt{2}}{2}+1 x=-\frac{\sqrt{2}}{2}+1$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$6\left(x-\left(\frac{\sqrt{2}}{2}+1\right)\right)\left(x-\left(-\frac{\sqrt{2}}{2}+1\right)\right)\geq 0$$

Koʻpaytma ≥0 boʻlishi uchun x-\left(\frac{\sqrt{2}}{2}+1\right) va x-\left(-\frac{\sqrt{2}}{2}+1\right) ikkalasi ≤0 yoki ≥0 boʻlishi kerak. x-\left(\frac{\sqrt{2}}{2}+1\right) va x-\left(-\frac{\sqrt{2}}{2}+1\right) ikkalasi ≤0 ga teng boʻlganda, yechimini toping.

$$x-\left(\frac{\sqrt{2}}{2}+1\right)\leq 0 x-\left(-\frac{\sqrt{2}}{2}+1\right)\leq 0$$

Ikkala tengsizlikning mos yechimi – x\leq -\frac{\sqrt{2}}{2}+1.

$$x\leq -\frac{\sqrt{2}}{2}+1$$

x-\left(\frac{\sqrt{2}}{2}+1\right) va x-\left(-\frac{\sqrt{2}}{2}+1\right) ikkalasi ≥0 ga teng boʻlganda, yechimini toping.

$$x-\left(-\frac{\sqrt{2}}{2}+1\right)\geq 0 x-\left(\frac{\sqrt{2}}{2}+1\right)\geq 0$$

Ikkala tengsizlikning mos yechimi – x\geq \frac{\sqrt{2}}{2}+1.

$$x\geq \frac{\sqrt{2}}{2}+1$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x\leq -\frac{\sqrt{2}}{2}+1\text{; }x\geq \frac{\sqrt{2}}{2}+1$$