Bài 1:
a: ĐKXĐ: \(\begin{cases}x-2\ge0\\ x-2<>0\end{cases}\Rightarrow x-2>0\)
=>x>2
b: ĐKXĐ: \(4x^2-4x+1\ge0\)
=>\(\left(2x-1\right)^2\ge0\) (luôn đúng)
=>x∈R
c: ĐKXĐ: \(x^2-x-6\ge0\)
=>(x-3)(x+2)>=0
=>x>=3 hoặc x<=-2
d: ĐKXĐ: \(\frac{-1}{x-1}\ge0\)
=>x-1<0
=>x<1
Bài 2:
a: \(\sqrt{\left(5-2\sqrt6\right)^2}-\sqrt{\left(5+2\sqrt6\right)}^2\)
\(=5-2\sqrt6-\left(5+2\sqrt6\right)\)
\(=5-2\sqrt6-5-2\sqrt6=-4\sqrt6\)
b: \(\sqrt{6-4\sqrt2}+\sqrt{22-12\sqrt2}\)
\(=\sqrt{\left(2-\sqrt2\right)^2}+\sqrt{18-2\cdot3\sqrt2\cdot2+4}\)
\(=2-\sqrt2+\sqrt{\left(3\sqrt2-2\right)^2}\)
\(=2-\sqrt2+3\sqrt2-2=2\sqrt2\)
c: \(\sqrt{5-\sqrt{13+4\sqrt3}}+\sqrt{3+\sqrt{13+4\sqrt3}}\)
\(=\sqrt{5-\sqrt{\left(2\sqrt3+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt3+1\right)^2}}\)
\(=\sqrt{5-2\sqrt3-1}+\sqrt{3+2\sqrt3+1}\)
\(=\sqrt{\left(\sqrt3-1\right)^2}+\sqrt{\left(\sqrt3+1\right)^2}=\sqrt3-1+\sqrt3+1=2\sqrt3\)
