(12/y)^2+5y^2=16 eching

$$( \frac { 12 } { y } ) ^ { 2 } + 5 y ^ { 2 } = 16$$

y uchun yechish

$$y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{-\frac{\arctan(\frac{\sqrt{41}}{2})i}{2}}}{5}\approx 1,866355157-1,372327065i$$

$$y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{-\arctan(\frac{\sqrt{41}}{2})i+2\pi i}{2}}}{5}\approx -1,866355157+1,372327065i$$

$$y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{\arctan(\frac{\sqrt{41}}{2})i}{2}}}{5}\approx 1,866355157+1,372327065i$$

$$y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{\arctan(\frac{\sqrt{41}}{2})i+2\pi i}{2}}}{5}\approx -1,866355157-1,372327065i$$

Kvadrat tenglamalaridan foydalanish qoidalari

\frac{12}{y}ni darajaga oshirish uchun, surat va maxrajni darajaga oshirib, keyin bo‘ling.

$$\frac{12^{2}}{y^{2}}+5y^{2}=16$$

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 5y^{2} ni \frac{y^{2}}{y^{2}} marotabaga ko'paytirish.

$$\frac{12^{2}}{y^{2}}+\frac{5y^{2}y^{2}}{y^{2}}=16$$

\frac{12^{2}}{y^{2}} va \frac{5y^{2}y^{2}}{y^{2}} da bir xil maxraji bor, ularning suratini qo‘shish orqali qo‘shing.

$$\frac{12^{2}+5y^{2}y^{2}}{y^{2}}=16$$

12^{2}+5y^{2}y^{2} ichidagi ko‘paytirishlarni bajaring.

$$\frac{12^{2}+5y^{4}}{y^{2}}=16$$

12^{2}+5y^{4} kabi iboralarga o‘xshab birlashtiring.

$$\frac{144+5y^{4}}{y^{2}}=16$$

Ikkala tarafdan 16 ni ayirish.

$$\frac{144+5y^{4}}{y^{2}}-16=0$$

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 16 ni \frac{y^{2}}{y^{2}} marotabaga ko'paytirish.

$$\frac{144+5y^{4}}{y^{2}}-\frac{16y^{2}}{y^{2}}=0$$

\frac{144+5y^{4}}{y^{2}} va \frac{16y^{2}}{y^{2}} da bir xil maxraji bor, ularning suratini ayirish orqali ayiring.

$$\frac{144+5y^{4}-16y^{2}}{y^{2}}=0$$

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

$$144+5y^{4}-16y^{2}=0$$

y^{2} uchun t ni almashtiring.

$$5t^{2}-16t+144=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 5 ni, b uchun -16 ni va c uchun 144 ni ayiring.

$$t=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 144}}{2\times 5}$$

Hisoblarni amalga oshiring.

$$t=\frac{16±\sqrt{-2624}}{10}$$

t=\frac{16±\sqrt{-2624}}{10} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{8+4\sqrt{41}i}{5} t=\frac{-4\sqrt{41}i+8}{5}$$

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

$$y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{\arctan(\frac{\sqrt{41}}{2})i+2\pi i}{2}}}{5} y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{\arctan(\frac{\sqrt{41}}{2})i}{2}}}{5} y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{-\frac{\arctan(\frac{\sqrt{41}}{2})i}{2}}}{5} y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{-\arctan(\frac{\sqrt{41}}{2})i+2\pi i}{2}}}{5}$$

y qiymati 0 teng bo‘lmaydi.

$$y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{-\arctan(\frac{\sqrt{41}}{2})i+2\pi i}{2}}}{5}\text{, }y\neq 0 y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{-\frac{\arctan(\frac{\sqrt{41}}{2})i}{2}}}{5}\text{, }y\neq 0 y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{\arctan(\frac{\sqrt{41}}{2})i}{2}}}{5}\text{, }y\neq 0 y=\frac{2\sqrt{3}\times 5^{\frac{3}{4}}e^{\frac{\arctan(\frac{\sqrt{41}}{2})i+2\pi i}{2}}}{5}\text{, }y\neq 0$$