\(x^2+4x=23-10\sqrt{2}\Leftrightarrow x^2+4x+4=\left(x+2\right)^2=27-10\sqrt{2}=25-10\sqrt{2}+2=5^2-5.2\sqrt{2}+\left(\sqrt{2}\right)^2=\left(5-\sqrt{2}\right)^2=\left(\sqrt{2}-5\right)^2\Leftrightarrow\left[{}\begin{matrix}x+2=5-\sqrt{2}\\x+2=\sqrt{2}-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3-\sqrt{2}\\x=\sqrt{2}-7\end{matrix}\right.\)
\(b,+,x>\frac{1}{2}\Rightarrow2x>1\Rightarrow1-2x< 0\Rightarrow\left|1-2x\right|=-\left(1-2x\right)=2x-1\Rightarrow\left(2x-1\right)^2=\left(2x-1\right)\Leftrightarrow\left(2x-1\right)\left(2x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(loai\right)\\x=1\left(thoaman\right)\end{matrix}\right.\)\(+,x\le\frac{1}{2}\Rightarrow2x\le1\Rightarrow1-2x\ge0\Rightarrow\left|1-2x\right|=1-2x\Rightarrow\left(2x-1\right)^2=1-2x\Leftrightarrow2x\left(2x-1\right)=0\Leftrightarrow x\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(thoaman\right)\\x=\frac{1}{2}\left(thoaman\right)\end{matrix}\right..\)
\(c,Taco:\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(2x+1\right)^2\ge0\end{matrix}\right.mà:\left(x-2\right)^2+\left(2x+1\right)^2=0nên:\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-\frac{1}{2}\end{matrix}\right.\left(voli\right).Nên:x\in\varnothing\)
