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

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

x uchun yechish

$$x=\frac{-\sqrt{29}-5}{2}\approx -5,192582404$$

$$x = \frac{\sqrt{29} + 5}{2} \approx 5,192582404$$

$$x=\frac{\sqrt{29}-5}{2}\approx 0,192582404$$

$$x=\frac{5-\sqrt{29}}{2}\approx -0,192582404$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

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

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

Ikkala tarafdan 27x^{2} ni ayirish.

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

x^{2} uchun t ni almashtiring.

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

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

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

Hisoblarni amalga oshiring.

$$t=\frac{27±5\sqrt{29}}{2}$$

t=\frac{27±5\sqrt{29}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{5\sqrt{29}+27}{2} t=\frac{27-5\sqrt{29}}{2}$$

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

$$x=\frac{\sqrt{29}+5}{2} x=-\frac{\sqrt{29}+5}{2} x=-\frac{5-\sqrt{29}}{2} x=\frac{5-\sqrt{29}}{2}$$