Bài 4 :
a, \(\sqrt{\left(2\sqrt{2}-3\right)^2}= \sqrt{\left(3-2\sqrt{2}\right)^2}=\left|3-2\sqrt{2}\right|=3-2\sqrt{2}\)
b, \(\sqrt{\left(\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\right)^2}=\left|\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\right|=\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}=\dfrac{1}{2}-\dfrac{\sqrt{2}}{2}=\dfrac{1-\sqrt{2}}{2}\)
c, \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}=\left|5-2\sqrt{6}\right|-\left|5+2\sqrt{6}\right|=-4\sqrt{6}\)
d, \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2\left(11-6\sqrt{2}\right)}\)
\(=2-\sqrt{2}+\sqrt{2\left(3-\sqrt{2}\right)^2}=2-\sqrt{2}+\sqrt{2}\left(3-\sqrt{2}\right)\)
\(=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
Bài 4:
a) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=3-2\sqrt{2}\)
b) \(\sqrt{\left(\dfrac{1}{2}-\dfrac{1}{\sqrt{2}}\right)^2}=\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}=\dfrac{\sqrt{2}-1}{2}\)
c) \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)
d) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
Bài 5:
a) Ta có: \(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\left|x-3\right|\)
\(=x+3+3-x\)
=6
b) Ta có: \(\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2+x=2x+2\)
c) Ta có: \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{x-1}{x-1}\)
=1
d) Ta có: \(\left|x-2\right|+\dfrac{\sqrt{x^2-4x+4}}{x-2}\)
\(=2-x+\dfrac{2-x}{x-2}\)
\(=2-x-\dfrac{2-x}{2-x}\)
\(=2-x-1=1-x\)
