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

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

x uchun yechish (complex solution)

$$x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i}{2}}\approx 1,169629851+0,931683417i$$

$$x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i+2\pi i}{2}}\approx -1,169629851-0,931683417i$$

$$x=\sqrt[4]{5}e^{-\frac{\arctan(\sqrt{19})i}{2}}\approx 1,169629851-0,931683417i$$

$$x=\sqrt[4]{5}e^{\frac{-\arctan(\sqrt{19})i+2\pi i}{2}}\approx -1,169629851+0,931683417i$$

Kvadrat tenglamalaridan foydalanish qoidalari

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

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

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

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

Ikkala tarafdan x^{2} ni ayirish.

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

x^{2} uchun t ni almashtiring.

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

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 -1 ni va c uchun 5 ni ayiring.

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

Hisoblarni amalga oshiring.

$$t=\frac{1±\sqrt{-19}}{2}$$

t=\frac{1±\sqrt{-19}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{1+\sqrt{19}i}{2} t=\frac{-\sqrt{19}i+1}{2}$$

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

$$x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i+2\pi i}{2}} x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i}{2}} x=\sqrt[4]{5}e^{-\frac{\arctan(\sqrt{19})i}{2}} x=\sqrt[4]{5}e^{\frac{-\arctan(\sqrt{19})i+2\pi i}{2}}$$

x qiymati 0 teng bo‘lmaydi.

$$x=\sqrt[4]{5}e^{\frac{-\arctan(\sqrt{19})i+2\pi i}{2}}\text{, }x\neq 0 x=\sqrt[4]{5}e^{-\frac{\arctan(\sqrt{19})i}{2}}\text{, }x\neq 0 x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i}{2}}\text{, }x\neq 0 x=\sqrt[4]{5}e^{\frac{\arctan(\sqrt{19})i+2\pi i}{2}}\text{, }x\neq 0$$