Câu 6: \(a=3\sqrt5=\sqrt{3^2\cdot5}=\sqrt{45};b=4\sqrt4=\sqrt{4^2\cdot4}=\sqrt{64}\)
\(c=\frac12\cdot\sqrt8=\sqrt{8\cdot\frac14}=\sqrt2;d=\sqrt{42}\)
mà 2<42<45<64
nên c<d<a<b
=>Chọn B
Câu 3: \(\sqrt{9x^2-6x+1}=5\)
=>\(\sqrt{\left(3x-1\right)^2}=5\)
=>|3x-1|=5
=>\(\left[\begin{array}{l}3x-1=5\\ 3x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=6\\ 3x=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-\frac43\end{array}\right.\)
=>Chọn A

giúp e với ạ=3






