Bài 3: ĐKXĐ: 4(3x-2)>=0
=>3x-2>=0
=>x>=2/3
\(\sqrt{4\left(3x-2\right)}=8\)
=>\(4\left(3x-2\right)=8^2=64\)
=>3x-2=16
=>3x=18
=>x=6(nhận)
Bài 2:
a: \(2\sqrt3=\sqrt{2^2\cdot3}=\sqrt{12};3\sqrt2=\sqrt{3^2\cdot2}=\sqrt{18}\)
mà 12<18
nên \(2\sqrt3<3\sqrt2\)
b: \(5+\sqrt7<5+\sqrt9=5+3=8\)

