Bài 2/
$A\,=\sqrt{9a^4}+3a^2-\sqrt{16a^6}\\\quad =\sqrt{(3a^2)^2}+3a^2-\sqrt{(4a^3)^2}\\\quad =|3a^2|+|3a^2|-|4a^3|\\\quad =3a^2+3a^2-|4a^3|(vì\,\,3a^2\ge 0)\\\quad =6a^2-4|a^3|$
Vậy $A=6a^2-4|a^3|$
$B\,=4x+1-\sqrt{4x^2-4x+1}\\\quad =4x+1-\sqrt{(2x-1)^2}\\\quad =4x+1-|2x-1|$
Vì $x\ge \dfrac 1 2$
$↔2x\ge 1\\↔2x-1\ge 0\\→|2x-1|=2x-1\\→B=4x+1-(2x-1)\\\quad\quad\,\,\,=4x+1-2x+1\\\quad\quad\,\,\, 2x+2$
Vậy $B=2x+2$
Bài 3/
a/ $\sqrt{(2-\sqrt 5)^2}-\sqrt 5\\=|2-\sqrt 5|-\sqrt 5\\=\sqrt 5-2-\sqrt 5(vì\,\,2-\sqrt 5<0)\\=-2$
Vậy $\sqrt{(2-\sqrt 5)^2}=-2$
b/ $\sqrt{(\sqrt 3-1)^2}+\sqrt{(-3)^2}-\sqrt{(\sqrt 3-2)^2}\\=|\sqrt 3-1|+|-3|-|\sqrt 3-2|\\=\sqrt 3-1+3-(2-\sqrt 3)(vì\,\,\sqrt 3-1>0,-3<0,\sqrt 3-2<0)\\=\sqrt 3+2-2+\sqrt 3\\=2\sqrt 3$
Vậy $\sqrt{(\sqrt 3-1)^2}+\sqrt{(-3)^2}-\sqrt{(\sqrt 3-2)^2}=2\sqrt 3$
Bài 4/ ĐK: $x\in\Bbb R$
$\sqrt{(x-2)^2}=3\\↔|x-2|=3\\↔\left[\begin{array}{1}x-2=3\\x-2=-3\end{array}\right.\\↔\left[\begin{array}{1}x=5\\x=-1\end{array}\right.$
Vậy $x∈\{5;-1\}$
