24*x^2+1/x^2=27 eching

$$24 \cdot x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = 27$$

x uchun yechish

$$x = -\frac{\sqrt{3 \sqrt{633} + 81}}{12} \approx -1,042427968$$

$$x = \frac{\sqrt{3 \sqrt{633} + 81}}{12} \approx 1,042427968$$

$$x=\frac{\sqrt{81-3\sqrt{633}}}{12}\approx 0,195816067$$

$$x=-\frac{\sqrt{81-3\sqrt{633}}}{12}\approx -0,195816067$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x^{2} ga ko'paytirish.

$$24x^{2}x^{2}+1=27x^{2}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 2 va 2 ni qo‘shib, 4 ni oling.

$$24x^{4}+1=27x^{2}$$

Ikkala tarafdan 27x^{2} ni ayirish.

$$24x^{4}+1-27x^{2}=0$$

x^{2} uchun t ni almashtiring.

$$24t^{2}-27t+1=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 24 ni, b uchun -27 ni va c uchun 1 ni ayiring.

$$t=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 24\times 1}}{2\times 24}$$

Hisoblarni amalga oshiring.

$$t=\frac{27±\sqrt{633}}{48}$$

t=\frac{27±\sqrt{633}}{48} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{\sqrt{633}}{48}+\frac{9}{16} t=-\frac{\sqrt{633}}{48}+\frac{9}{16}$$

x=t^{2} boʻlganda, yechimlar har bir t uchun x=±\sqrt{t} hisoblanishi orqali olinadi.

$$x=\frac{\sqrt{\frac{\sqrt{633}}{3}+9}}{4} x=-\frac{\sqrt{\frac{\sqrt{633}}{3}+9}}{4} x=\frac{\sqrt{-\frac{\sqrt{633}}{3}+9}}{4} x=-\frac{\sqrt{-\frac{\sqrt{633}}{3}+9}}{4}$$