a: ĐKXĐ: x>=3
Ta có: \(\sqrt{9x-27}-2\cdot\sqrt{\frac{x-3}{4}}=5-2\cdot\sqrt{16x-48}\)
=>\(3\cdot\sqrt{x-3}-2\cdot\frac{\sqrt{x-3}}{2}+2\cdot4\sqrt{x-3}=5\)
=>\(3\cdot\sqrt{x-3}-\sqrt{x-3}+8\sqrt{x-3}=5\)
=>\(10\cdot\sqrt{x-3}=5\)
=>\(\sqrt{x-3}=\frac12\)
=>\(x-3=\frac14\)
=>\(x=3+\frac14=\frac{13}{4}\) (nhận)
b; \(\sqrt{4x^2-4x+1}-5=6x\)
=>\(\sqrt{\left(2x-1\right)^2}=6x+5\)
=>|2x-1|=6x+5
=>\(\begin{cases}6x+5\ge0\\ \left(6x+5\right)^2=\left(2x-1\right)^2\end{cases}\Rightarrow\begin{cases}6x\ge-5\\ \left(6x+5-2x+1\right)\left(6x+5+2x-1\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-\frac56\\ \left(4x+6\right)\left(8x+4\right)=0\end{cases}\)
=>x=-1/2
