sqrt[4x]{y^2+3}=x^8x^2 шешу

$$\sqrt[ 4x ]{ { y }^{ 2 } +3 } = { x }^{ 8 { x }^{ 2 } }$$

y мәнін табыңыз (complex solution)

$$y=e^{4\left(Im(\frac{1}{x})+iRe(\frac{1}{x})\right)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)arg(x^{8x^{2}})-8\pi Re(\frac{1}{x})n_{1}i\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)-8\pi Im(\frac{1}{x})n_{1}\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}\left(|x^{8x^{2}}|\right)^{4\left(Re(\frac{1}{x})-iIm(\frac{1}{x})\right)\left(\left(Re(x)\right)^{2}+\left(Im(x)\right)^{2}\right)}$$

$$n_{1}\in \mathrm{Z}$$

$$x\neq 0$$

y мәнін табыңыз

$$y=\sqrt{\left(x^{8x^{2}}\right)^{4x}-3}$$

$$y=-\sqrt{\left(x^{8x^{2}}\right)^{4x}-3}\text{, }\left(x>0\text{ and }x^{8x^{3}}\geq \sqrt[4]{3}\right)\text{ or }\left(x<0\text{ and }\left(-x\right)^{8x^{3}}\geq \sqrt[4]{3}\text{ and }Denominator(x^{2})\text{bmod}2=1\right)$$