3) \(\sqrt{\left(x-2\right)\left(x+1\right)}\) thì (x-2)(x+1)>0
=> x2 -x-2>0
=> x2 - x - \(\dfrac{1}{2}\)- \(\dfrac{3}{2}\)>0
= (x+\(\dfrac{1}{4}\))2 - 3/2 >0
=> x+ 1/4>3/2
=> x>5/4
4) Có x đâu mà tìm bạn??
4) \(\sqrt{x^2+2x+1}\) + \(\sqrt{x^2-2x+1}\)= \(\sqrt{\left(x+1\right)^2}\) + \(\sqrt{\left(x-1\right)^2}\)
=> /x+1/+/x-1/ = 2
=> /2x/ = 2
=> 2x=2
=> x=1
Để biểu thức có nghĩa thì (x-2)(x+1)>0
\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)