a đk x>= 3
, \(\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\\sqrt{x+3}+\sqrt{x-3}=0\end{matrix}\right.\)
Vì \(\sqrt{x+3};\sqrt{x-3}\ge0\Rightarrow\left\{{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\left(ktm\right)\)
Vậy x = 9
b, \(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)
Với x < 1 => 1 - x + 2 - x = 3 <=> -2x = 0 <=> x = 0 (tm)
Với x >= 2 <=> x - 1 + x - 2 = 3 <=> 2x - 3 = 3 <=> x = 3 (tm)
Với 1 =< x < 2 <=> x - 1 + 2 - x =3 <=> 1 = 3 (vô lí)
