giúp mik vs
|HH   BVN   BƯN,Bàn C ,\Leftrightarrow\frac{|x|^{2}-|x|}{|x|-|^{2}x|},Va   \begin{cases}{\int_{-2}^{-1}(-f(x)+(\frac{-\sqrt{x}}{x+\sqrt{x}+1})-(\frac{-\sqrt{x}}{x+\sqrt{x+\sqrt{x}+1})=-2}}\ {|\sqrt{x+\sqrt{x}+1}|-|\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\frac{+}}+1}}^{2}}}}}}\ \end{cases}.,\int_{-1}^{4}\sqrt{|x|-|x|^{4}+|x|^{2}}+|\sqrt{|x|^{4}+|x|^{2}}|+|\sqrt{|x|^{2}}|^{4},7o,\begin{cases}{\frac{1}{x+1}+\sqrt{1+\sqrt{x+1}+1}+(\sqrt{1+\sqrt{x+1}+1}+1)-(\sqrt{x+2}+1)+(\sqrt{x+2}+1)+(\sqrt{x+1}+1)-(\sqrt{x+x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+\sqrt{y})+\sqrt{x+1}+(\sqrt{y})+(\sqrt{y})+1)}\end{cases}.,x ,\frac{+\begin{matrix}{{-+\sqrt{x}=\frac{-+\sqrt{x}}{x}=\frac{-\sqrt{x}}{x}+\frac{\sqrt{x}+\sqrt{y}}{x}=\sqrt{x}-\sqrt{y}}\ {\mathtt{+\sqrt{x}+\sqrt{y}=\sqrt{x}-\sqrt{y}}\ \end{matrix}}\geq\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\end{matrix}}\ ,

Question

giúp mik vs
 |HH &nbsp; BVN          &nbsp; BƯN,Bàn    C     ,\Leftrightarrow\frac{|x|^{2}-|x|}{|x|-|^{2}x|},Va &nbsp; \begin{cases}{\int_{-2}^{-1}(-f(x)+(\frac{-\sqrt{x}}{x+\sqrt{x}+1})-(\frac{-\sqrt{x}}{x+\sqrt{x+\sqrt{x}+1})=-2}}\ {|\sqrt{x+\sqrt{x}+1}|-|\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\frac{+}}+1}}^{2}}}}}}\ \end{cases}.,\int_{-1}^{4}\sqrt{|x|-|x|^{4}+|x|^{2}}+|\sqrt{|x|^{4}+|x|^{2}}|+|\sqrt{|x|^{2}}|^{4},7o,\begin{cases}{\frac{1}{x+1}+\sqrt{1+\sqrt{x+1}+1}+(\sqrt{1+\sqrt{x+1}+1}+1)-(\sqrt{x+2}+1)+(\sqrt{x+2}+1)+(\sqrt{x+1}+1)-(\sqrt{x+x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+1)+(\sqrt{x+1}+\sqrt{y})+\sqrt{x+1}+(\sqrt{y})+(\sqrt{y})+1)}\end{cases}.,x ,\frac{+\begin{matrix}{{-+\sqrt{x}=\frac{-+\sqrt{x}}{x}=\frac{-\sqrt{x}}{x}+\frac{\sqrt{x}+\sqrt{y}}{x}=\sqrt{x}-\sqrt{y}}\ {\mathtt{+\sqrt{x}+\sqrt{y}=\sqrt{x}-\sqrt{y}}\ \end{matrix}}\geq\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\end{matrix}}\ ,

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Timi · Answer

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