0\\-x>0\end{cases}\Rightarrow\begin{cases}x(4+x)<0\\x<0\end{cases}\Rightarrow x\in(-4;\;0)\\\log_{\sqrt3}(-4x-x^2)-\log_{3^\frac13}(-x)=2\\\log_{\sqrt3}(-4x-x^2)-3\log_{\sqrt3}(-x)=2\\\log_{\sqrt3}(-4x-x^2)-\log_{\sqrt3}(-x)^3=2\\\log_{\sqrt3}\frac{-4x-x^2}{-x^3}=2\\\frac{4+x}{x^2}=(\sqrt3)^2\\\frac{4+x}{x^2}=3\\3x^2-x-4=0" alt="1.\;\log_{\sqrt3}(-4x-x^2)-\log_{\sqrt[3]3}(-x)=2\\O.D.3.:\\\begin{cases}-4x-x^2>0\\-x>0\end{cases}\Rightarrow\begin{cases}x(4+x)<0\\x<0\end{cases}\Rightarrow x\in(-4;\;0)\\\log_{\sqrt3}(-4x-x^2)-\log_{3^\frac13}(-x)=2\\\log_{\sqrt3}(-4x-x^2)-3\log_{\sqrt3}(-x)=2\\\log_{\sqrt3}(-4x-x^2)-\log_{\sqrt3}(-x)^3=2\\\log_{\sqrt3}\frac{-4x-x^2}{-x^3}=2\\\frac{4+x}{x^2}=(\sqrt3)^2\\\frac{4+x}{x^2}=3\\3x^2-x-4=0" align="absmiddle" class="latex-formula">
Ответ: x = -1
![2.\;\frac{x^2-8y^2}{x^\frac23-2\sqrt[3]{y^2}}-\left(x^\frac23+2\sqrt[3]{y^2}\right)^2=-2x^\frac23y^\frac23\\\frac{x^2-8y^2}{x^\frac23-2\sqrt[3]{y^2}}-\left(x^\frac23+2\sqrt[3]{y^2}\right)^2=\frac{x^2-8y^2}{x^\frac23-2{y^\frac23}}-\left(x^\frac23+2y^\frac23\right)^2=\\=\frac{\left(x^\frac23-2y^\frac23\right)\left(x^\frac43+2x^\frac23y^\frac23+4y^\frac43\right)}{x^\frac23-2{y^\frac23}}-\left(x^\frac43+4x^\frac23y^\frac23+4y^\frac43\right)=](https://tex.z-dn.net/?f=2.%5C%3B%5Cfrac%7Bx%5E2-8y%5E2%7D%7Bx%5E%5Cfrac23-2%5Csqrt%5B3%5D%7By%5E2%7D%7D-%5Cleft%28x%5E%5Cfrac23%2B2%5Csqrt%5B3%5D%7By%5E2%7D%5Cright%29%5E2%3D-2x%5E%5Cfrac23y%5E%5Cfrac23%5C%5C%5Cfrac%7Bx%5E2-8y%5E2%7D%7Bx%5E%5Cfrac23-2%5Csqrt%5B3%5D%7By%5E2%7D%7D-%5Cleft%28x%5E%5Cfrac23%2B2%5Csqrt%5B3%5D%7By%5E2%7D%5Cright%29%5E2%3D%5Cfrac%7Bx%5E2-8y%5E2%7D%7Bx%5E%5Cfrac23-2%7By%5E%5Cfrac23%7D%7D-%5Cleft%28x%5E%5Cfrac23%2B2y%5E%5Cfrac23%5Cright%29%5E2%3D%5C%5C%3D%5Cfrac%7B%5Cleft%28x%5E%5Cfrac23-2y%5E%5Cfrac23%5Cright%29%5Cleft%28x%5E%5Cfrac43%2B2x%5E%5Cfrac23y%5E%5Cfrac23%2B4y%5E%5Cfrac43%5Cright%29%7D%7Bx%5E%5Cfrac23-2%7By%5E%5Cfrac23%7D%7D-%5Cleft%28x%5E%5Cfrac43%2B4x%5E%5Cfrac23y%5E%5Cfrac23%2B4y%5E%5Cfrac43%5Cright%29%3D "2.\;\frac{x^2-8y^2}{x^\frac23-2\sqrt[3]{y^2}}-\left(x^\frac23+2\sqrt[3]{y^2}\right)^2=-2x^\frac23y^\frac23\\\frac{x^2-8y^2}{x^\frac23-2\sqrt[3]{y^2}}-\left(x^\frac23+2\sqrt[3]{y^2}\right)^2=\frac{x^2-8y^2}{x^\frac23-2{y^\frac23}}-\left(x^\frac23+2y^\frac23\right)^2=\\=\frac{\left(x^\frac23-2y^\frac23\right)\left(x^\frac43+2x^\frac23y^\frac23+4y^\frac43\right)}{x^\frac23-2{y^\frac23}}-\left(x^\frac43+4x^\frac23y^\frac23+4y^\frac43\right)=")
ЧТД