a: ĐKXĐ: x>=-3/5
Ta có: \(3\cdot\sqrt{20x+12}-\frac23\cdot\sqrt{45x+27}=2\)
=>\(3\cdot2\cdot\sqrt{5x+3}-\frac23\cdot3\cdot\sqrt{5x+3}=2\)
=>\(4\cdot\sqrt{5x+3}=2\)
=>\(\sqrt{5x+3}=\frac12\)
=>\(5x+3=\frac14\)
=>\(5x=\frac14-3=-\frac{11}{4}\)
=>\(x=-\frac{11}{4}:5=-\frac{11}{20}\) (nhận)
b: ĐKXĐ: x>=3
\(\sqrt{4x-12}-6\cdot\sqrt{\frac{x-3}{4}}=10-\sqrt{9x-27}\)
=>\(2\sqrt{x-3}-3\cdot\sqrt{x-3}+3\sqrt{x-3}=10\)
=>\(2\cdot\sqrt{x-3}=10\)
=>\(\sqrt{x-3}=5\)
=>x-3=25
=>x=25+3=28(nhận)
c: \(\sqrt{x^2+9}+3=2x\)
=>\(\sqrt{x^2+9}=2x-3\)
=>\(\begin{cases}2x-3\ge0\\ \left(2x-3\right)^2=x^2+9\end{cases}\Rightarrow\begin{cases}x\ge\frac32\\ 4x^2-12x+9-x^2-9=0\end{cases}\)
=>\(\begin{cases}x\ge\frac32\\ 3x^2-12x=0\end{cases}\Rightarrow\begin{cases}x\ge\frac32\\ 3x\left(x-4\right)=0\end{cases}\Rightarrow x=4\)
d: \(\sqrt{9x^2-6x+1}+5x=1\)
=>\(\sqrt{\left(3x-1\right)^2}+5x=1\)
=>|3x-1|=1-5x
=>\(\begin{cases}1-5x\ge0\\ \left(1-5x\right)^2=\left(3x-1\right)^2\end{cases}\Rightarrow\begin{cases}5x\le1\\ \left(5x-1-3x+1\right)\left(5x-1+3x-1\right)=0\end{cases}\)
=>\(\begin{cases}x\le\frac15\\ 2x\cdot\left(8x-2\right)=0\end{cases}\Rightarrow x=0\)
