0+frac{2m/m^2+frac{4{3}}}{x+2m^2/m^2+frac{4{3}}+2}=2/m-frac{2m/m^2+frac{4{3}}}{frac{2{m}^2}{{m}^2+frac{4}{3}}} шешу

$$\frac{ 0+ \frac{ 2m }{ { m }^{ 2 } + \frac{ 4 }{ 3 } } }{ x+ \frac{ 2 { m }^{ 2 } }{ { m }^{ 2 } + \frac{ 4 }{ 3 } } +2 } = \frac{ \frac{ 2 }{ m } - \frac{ 2m }{ { m }^{ 2 } + \frac{ 4 }{ 3 } } }{ \frac{ 2 { m }^{ 2 } }{ { m }^{ 2 } + \frac{ 4 }{ 3 } } }$$

m мәнін табыңыз

$$\left\{\begin{matrix}m=-\frac{\sqrt{3\left(-\sqrt{x^{2}+16x+32}+x+4\right)}}{3}\text{, }&-\sqrt{\frac{-\sqrt{x^{2}+16x+32}+x+4}{3}}\neq -\frac{2\sqrt{-3x-6}}{3\sqrt{x+4}}\text{ and }x\geq 4\sqrt{2}-8\text{ and }x<-2\\m=\frac{\sqrt{3\left(-\sqrt{x^{2}+16x+32}+x+4\right)}}{3}\text{, }&\sqrt{\frac{-\sqrt{x^{2}+16x+32}+x+4}{3}}\neq \frac{2\sqrt{-3x-6}}{3\sqrt{x+4}}\text{ and }x\geq 4\sqrt{2}-8\text{ and }x<-2\\m=-\frac{\sqrt{3\left(\sqrt{x^{2}+16x+32}+x+4\right)}}{3}\text{, }&x\geq 4\sqrt{2}-8\text{ and }\left(-\sqrt{\frac{\sqrt{x^{2}+16x+32}+x+4}{3}}\neq -\frac{2\sqrt{-3x-6}}{3\sqrt{x+4}}\text{ or }x>-2\right)\\m=\frac{\sqrt{3\left(\sqrt{x^{2}+16x+32}+x+4\right)}}{3}\text{, }&x\geq 4\sqrt{2}-8\text{ and }\left(\sqrt{\frac{\sqrt{x^{2}+16x+32}+x+4}{3}}\neq \frac{2\sqrt{-3x-6}}{3\sqrt{x+4}}\text{ or }x>-2\right)\end{matrix}\right.$$

x мәнін табыңыз

$$x=-\frac{16+24m^{2}-9m^{4}}{2\left(3m^{2}+4\right)}$$

$$m\neq 0$$