$6.\\a)\sqrt{(3-\sqrt{5})^2}=|3-\sqrt{5}|=3-\sqrt{5}\\
b)\sqrt{(5-\sqrt{11})^2}=|5-\sqrt{11}|=5-\sqrt{11}\\
c)\sqrt{(\sqrt{10}-3)^2}=|\sqrt{10}-3|=\sqrt{10}-3\\
d)\sqrt{(2-\sqrt{5})^2}=|2-\sqrt{5}|=\sqrt{5}-2\\
e)\sqrt{(\sqrt{13}-5)^2}=|\sqrt{13}-5|=5-\sqrt{13}\\
f)\sqrt{(\sqrt{2}+3)^2}=|\sqrt{2}+3|=3+\sqrt{2}\\
g)\sqrt{(4+\sqrt{3})^2}=|4+\sqrt{3}|=4+\sqrt{3}\\
h)\sqrt{(4-\sqrt{17})^2}=|4-\sqrt{17}|=\sqrt{17}-4\\
i)\sqrt{(4-2\sqrt{7})^2}=|4-2\sqrt{7}|=2\sqrt7-4$
$7.$
$a)\sqrt{x^2}=|x|=x$ (do $x\geq 0$)
$b)\sqrt{x^2}=|x|=-x$ (do $x<0$)
$c)2\sqrt{x^2}=2|x|=-2x$ (do $x<0$)
$d)-\dfrac{1}{3}\sqrt{9x^2}=-\dfrac13.3|x|=-|x|=-x$ (do $x\geq 0$)
$e)\sqrt{(a-3)^2}=|a-3|=\left\{\begin{matrix}a-3 \ \text{nếu} \ a\geq 3\\ 3-a \ \text{nếu} \ 0\leq a<3\end{matrix}\right.$
$f)\sqrt{(x-5)^2}=|x-5|=5-x$ (do $x<5\Leftrightarrow x-5<0$)
$g)2\sqrt{(x-2)^2}=2|x-2|=2(2-x)=4-2x$ (do $x<2\Leftrightarrow x-2<0$)
$h)\sqrt{x^2+6x+9}=\sqrt{(x+3)^2}=|x+3|=x+3$ (do $x\geq -3\Leftrightarrow x+3\geq 0$)
$i)\sqrt{(a-2b)^2}=|a-2b|=2b-a$ (do $a\leq 2b\Leftrightarrow a-2b\leq 0$)
$j)\sqrt{4x^2-4xy+y^2}=\sqrt{(2x-y)^2}=|2x-y|=2x-y$ (do $2x\geq y\Leftrightarrow 2x-y\geq 0$)
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
