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a) \(\left|x^2+4\right|=4x\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+4=4x\\x^2+4=-4x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-4x+4=0\\x^2+4x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(x+2\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(TMĐK\right)\\x=-2\left(KTMĐK\right)\end{cases}}\)

Vậy \(S=\left\{2\right\}\)

b) \(\left|2-4x\right|=2x+1\left(ĐK:x\ge\frac{1}{2}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}2-4x=2x+1\\2-4x=-2x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-4x-2x=-2+1\\-4x+2x=-2-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-6x=-1\\-2x=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{6}\left(KTMĐK\right)\\x=\frac{3}{2}\left(TMĐK\right)\end{cases}}\)

Vậy \(S=\left\{\frac{3}{2}\right\}\)

ta có : \(y^4=x^2.z^2=y.z.z^2\Leftrightarrow\orbr{\begin{cases}y=0\\y=z\end{cases}\Rightarrow y=z\Rightarrow x=y=z}\)

Vậy \(A=\frac{\left(3x\right)^{2022}}{x^{2022}}=3^{2022}\)

\(\frac{\left(-3\right)^2}{x+1}=\frac{9}{x+1}\left(ĐKXĐ:x\ne1\right)\)

Vì \(9>0\Rightarrow\)Để có số hứu tỉ âm thì \(x+1< 0\Leftrightarrow x< -1\)

\(\left|x-1,5\right|+\left|2,5-x\right|=0\)

Ta có:

\(\hept{\begin{cases}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left|x-1,5\right|=0\\\left|2,5-x\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-1,5=0\\2,5-x=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1,5\\x=2,5\end{cases}}\) (Vô lý)

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