Bài 1:
a: \(-0,8\cdot\sqrt{\left(-0,125\right)^2}=-0,8\cdot\left|-0,125\right|\)
\(=-0,8\cdot0,125=-1\)
b: \(\sqrt{\left(-2\right)^6}=\sqrt{64}=8\)
c: \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)
d: \(\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|3-2\sqrt{2}\right|=3-2\sqrt{2}\)
f: \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}=\left|0,1-\sqrt{0,1}\right|=\left|\dfrac{1}{10}-\dfrac{1}{\sqrt{10}}\right|\)
\(=\left|\dfrac{1-\sqrt{10}}{10}\right|=\dfrac{\sqrt{10}-1}{10}\)
e: \(\sqrt{\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right)^2}=\left|\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right|=\left|\dfrac{\sqrt{2}-1}{2}\right|=\dfrac{\sqrt{2}-1}{2}\)
Bài 2:
a: \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|+\left|3+2\sqrt{2}\right|\)
\(=3-2\sqrt{2}+3+2\sqrt{2}=6\)
b: \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=\left|5-2\sqrt{6}\right|-\left|5+2\sqrt{6}\right|\)
\(=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)
c: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(=2-\sqrt{3}+\sqrt{3}-1=1\)
d: \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|-\left|1-\sqrt{2}\right|\)
\(=3+\sqrt{2}-\sqrt{2}+1=4\)
