x^3+1/x^3=3 eching

$${ x }^{ 3 } + \frac{ 1 }{ { x }^{ 3 } } =3$$

x uchun yechish (complex solution)

$$x\in \frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+3}e^{\frac{2\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+3}e^{\frac{4\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{5}+3}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}e^{\frac{4\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}e^{\frac{2\pi i}{3}}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x^{3} ga ko'paytirish.

$$x^{3}x^{3}+1=3x^{3}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 3 va 3 ni qo‘shib, 6 ni oling.

$$x^{6}+1=3x^{3}$$

Ikkala tarafdan 3x^{3} ni ayirish.

$$x^{6}+1-3x^{3}=0$$

x^{3} uchun t ni almashtiring.

$$t^{2}-3t+1=0$$

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 -3 ni va c uchun 1 ni ayiring.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Hisoblarni amalga oshiring.

$$t=\frac{3±\sqrt{5}}{2}$$

t=\frac{3±\sqrt{5}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{\sqrt{5}+3}{2} t=\frac{3-\sqrt{5}}{2}$$

x=t^{3} boʻlganda, yechimlar har bir t uchun tenglamani yechish orqali olinadi.

$$x=-\sqrt[3]{\frac{\sqrt{5}+3}{2}}e^{\frac{\pi i}{3}} x=\sqrt[3]{\frac{\sqrt{5}+3}{2}}ie^{\frac{\pi i}{6}} x=\sqrt[3]{\frac{\sqrt{5}+3}{2}} x=-\sqrt[3]{\frac{3-\sqrt{5}}{2}}e^{\frac{\pi i}{3}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}ie^{\frac{\pi i}{6}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}$$

x qiymati 0 teng bo‘lmaydi.

$$x=\sqrt[3]{\frac{3-\sqrt{5}}{2}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}ie^{\frac{\pi i}{6}}\text{, }x\neq 0 x=-\sqrt[3]{\frac{3-\sqrt{5}}{2}}e^{\frac{\pi i}{3}}\text{, }x\neq 0 x=\sqrt[3]{\frac{\sqrt{5}+3}{2}} x=\sqrt[3]{\frac{\sqrt{5}+3}{2}}ie^{\frac{\pi i}{6}}\text{, }x\neq 0 x=-\sqrt[3]{\frac{\sqrt{5}+3}{2}}e^{\frac{\pi i}{3}}\text{, }x\neq 0$$

x uchun yechish

$$x=\frac{2^{\frac{2}{3}}\sqrt[3]{3-\sqrt{5}}}{2}\approx 0,72556263$$

$$x = \frac{2 ^ {\frac{2}{3}} \sqrt[3]{\sqrt{5} + 3}}{2} \approx 1,378240772$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x^{3} ga ko'paytirish.

$$x^{3}x^{3}+1=3x^{3}$$

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 3 va 3 ni qo‘shib, 6 ni oling.

$$x^{6}+1=3x^{3}$$

Ikkala tarafdan 3x^{3} ni ayirish.

$$x^{6}+1-3x^{3}=0$$

x^{3} uchun t ni almashtiring.

$$t^{2}-3t+1=0$$

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 -3 ni va c uchun 1 ni ayiring.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Hisoblarni amalga oshiring.

$$t=\frac{3±\sqrt{5}}{2}$$

t=\frac{3±\sqrt{5}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{\sqrt{5}+3}{2} t=\frac{3-\sqrt{5}}{2}$$

x=t^{3} boʻlganda, yechimlar har bir t uchun x=\sqrt[3]{t} hisoblanishi orqali olinadi.

$$x=\sqrt[3]{\frac{\sqrt{5}+3}{2}} x=\sqrt[3]{\frac{3-\sqrt{5}}{2}}$$