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)

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

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

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

Ta có: \(\log_{^{}\frac15}^{}x^2=\frac{\log_5x^2}{\log_5\frac15}=\frac{2\cdot\log_5x}{-1}=-2\cdot\log_5x\)

=>\(\left(\log_{\frac15}x^2\right)^2=4\cdot\log_5^2x\)

Ta có: \(\log_{\frac15}^2x^2+\log_5x+2=0\)

=>\(4\cdot\log_5^2x+\log_5x+2=0\)

Đặt \(a=\log_5x\)

Phương trình sẽ trở thành: \(4a^2+a+2=0\)

\(\Delta=1^2-4\cdot4\cdot2=1-32=-31<0\)

=>Phương trình vô nghiệm

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