a: \(\sqrt{9x^2-6x+1}=x-2\)
=>\(\sqrt{\left(3x-1\right)^2}=x-2\)
=>|3x-1|=x-2
=>\(\begin{cases}x-2\ge0\\ \left(3x-1\right)^2=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge2\\ \left(3x-1-x+2\right)\left(3x-1+x-2\right)=0\end{cases}\)
=>\(\begin{cases}x\ge2\\ \left(2x+1\right)\left(4x-3\right)=0\end{cases}\)
=>x∈∅
b: ĐKXĐ: \(x^2+4x\ge0\)
=>x(x+4)>=0
=>x>=0 hoặc x<=-4
\(\sqrt{x^2+4x}=2x-1\)
=>\(\begin{cases}2x-1\ge0\\ \left(2x-1\right)^2=x^2+4x\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ 4x^2-4x+1-x^2-4x=0\end{cases}\)
=>\(\begin{cases}x\ge\frac12\\ 3x^2-8x+1=0\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ x^2-\frac83x+\frac13=0\end{cases}\)
=>\(\begin{cases}x\ge\frac12\\ x^2-2\cdot x\cdot\frac43+\frac{16}{9}-\frac{13}{9}=0\end{cases}\)
=>\(\begin{cases}x\ge\frac12\\ \left.\left(x-\frac43\right)^2\right.=\frac{13}{9}\end{cases}\Rightarrow\) \(x=\frac{4+\sqrt{13}}{3}\)
c: ĐKXĐ: x>=2
\(3\cdot\sqrt{4x-8}-10=2\cdot\sqrt{\frac{x-2}{4}}\)
=>\(3\cdot2\sqrt{x-2}-10=2\cdot\frac{\sqrt{x-2}}{2}=\sqrt{x-2}\)
=>\(5\cdot\sqrt{x-2}=10\)
=>\(\sqrt{x-2}=2\)
=>x-2=4
=>x=6(nhận)
