4cm2

a) \(x+\sqrt{4x^2-4x+1}=2\)

\(\Leftrightarrow x+\sqrt{\left(2x-1\right)^2}=2\)

\(\Leftrightarrow x+|2x-1|=2\)

\(TH1:x\ge0\)

\(\Leftrightarrow x+2x-1=2\)

\(\Leftrightarrow3x-1=2\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\left(TM\right)\)

\(TH2:x< 0\)

\(\Leftrightarrow x-2x-1=2\)

\(\Leftrightarrow-x-1=2\)

\(\Leftrightarrow-x=3\)

\(\Leftrightarrow x=-3\left(TM\right)\)

Vậy:...

b) \(3x-1-\sqrt{4x^2-12x+9}=0\)

\(\Leftrightarrow3x-1-\sqrt{\left(2x-3\right)^2}=0\)

\(\Leftrightarrow3x-1-|2x-3|=0\)

\(TH1:x\ge0\)

\(\Leftrightarrow3x-1-2x+3=0\)

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(KTM\right)\)

\(TH2:x< 0\)

\(\Leftrightarrow3x-1+2x-3=0\)

\(\Leftrightarrow5x-4=0\Leftrightarrow x=\frac{4}{5}\left(KTM\right)\)

Vậy: pt vô nghiệm

Học Tốt!!!

a) \(\sqrt{a}+1>\sqrt{a+1}\)\(\Leftrightarrow\)\(a+2\sqrt{a}+1>a+1\)\(\Leftrightarrow\)\(2\sqrt{a}>0\)( luôn đúng \(\forall x>0\) ) 

b) \(a-1< a\)\(\Leftrightarrow\)\(\sqrt{a-1}< \sqrt{a}\)

c) \(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1>3-2\sqrt{3.2}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)

do \(\sqrt{6}-1>0;\sqrt{3}-\sqrt{2}>0\) nên \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\) ( đpcm )