x^2+1/x^2=5 eching

$${ x }^{ 2 } + \frac{ 1 }{ { x }^{ 2 } } =5$$

x uchun yechish

$$x=\frac{-\sqrt{3}-\sqrt{7}}{2}\approx -2,188901059$$

$$x = \frac{\sqrt{3} + \sqrt{7}}{2} \approx 2,188901059$$

$$x=\frac{\sqrt{7}-\sqrt{3}}{2}\approx 0,456850252$$

$$x=\frac{\sqrt{3}-\sqrt{7}}{2}\approx -0,456850252$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

$$x^{2}x^{2}+1=5x^{2}$$

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

$$x^{4}+1=5x^{2}$$

Ikkala tarafdan 5x^{2} ni ayirish.

$$x^{4}+1-5x^{2}=0$$

x^{2} uchun t ni almashtiring.

$$t^{2}-5t+1=0$$

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 -5 ni va c uchun 1 ni ayiring.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 1}}{2}$$

Hisoblarni amalga oshiring.

$$t=\frac{5±\sqrt{21}}{2}$$

t=\frac{5±\sqrt{21}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{\sqrt{21}+5}{2} t=\frac{5-\sqrt{21}}{2}$$

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

$$x=\frac{\sqrt{3}+\sqrt{7}}{2} x=-\frac{\sqrt{3}+\sqrt{7}}{2} x=-\frac{\sqrt{3}-\sqrt{7}}{2} x=\frac{\sqrt{3}-\sqrt{7}}{2}$$