Question

Answer 1

a, đk \(x\ge\dfrac{1}{3}\)

\(\left(\sqrt{3x-1}\right)^2=\left(\sqrt{5+x}\right)^2\\ 3x-1=5+x\\ 3x-x=5+1\\ 2x=6\\ x=3\left(thoaman\right)\\ b,đkx\ge\dfrac{1}{2}\\ \sqrt{\left(2x-1\right)^2}=3+x\\ \left|2x-1\right|=3+x\\ 2x-1-x=3\\ x=4\left(thoaman\right)\)

Answer 2

a, x >= 1/3 

<=> 3x - 1 = x + 5 <=> 2x = 6 <=> x = 3 (tm) 

b, \(\sqrt{\left(2x-1\right)^2}=x+3\)đk x >= -3 

\(\left[{}\begin{matrix}2x-1=x+3\\2x-1=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)(tm)