Long Vũ Hoàng

doi ti

\(\left|x\left(x^x-\frac{5}{4}\right)\right|=0\)

\(\Rightarrow x\left(x^x-\frac{5}{4}\right)=0\)

\(\Rightarrow x=0\) hoặc \(x^x-\frac{5}{4}=0\)

\(\Rightarrow x=0\) hoặc \(x^x=\frac{5}{4}\Rightarrow x=\frac{\sqrt{5}}{2}\) 

Vậy \(x\in\left\{0;\frac{\sqrt{5}}{2}\right\}\)

\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

\(=>\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)

\(=>\left(3x+2\right)\left(x-4\right)=0\)

\(=>\left[{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}3x=-2\\x=4\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=4\end{matrix}\right.\)

\(=>x\in\left\{\dfrac{-2}{3};4\right\}\)

\(\left(2x-1\right)^2-\left(x+3\right)^2=0\)(sửa đề)

\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\)           \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)