1: ĐKXĐ: x>=1
Ta có: \(\sqrt{3x+1}+\sqrt{x-1}=2\)
=>\(\sqrt{3x+1}-2+\sqrt{x-1}=0\)
=>\(\frac{3x+1-4}{\sqrt{3x+1}+2}+\sqrt{x-1}=0\)
=>\(\frac{3x-3}{\sqrt{3x-1}+2}+\sqrt{x-1}=0\)
=>\(\sqrt{x-1}\left(\frac{3\sqrt{x-1}}{\sqrt{3x-1}+2}+1\right)=0\)
=>\(\sqrt{x-1}=0\)
=>x-1=0
=>x=1(nhận)
2: ĐKXĐ: x>=1/5
Ta có: \(\sqrt{5x-1}-\sqrt{x}=4x-1\)
=>\(\frac{5x-1-x}{\sqrt{5x-1}+\sqrt{x}}=4x-1\)
=>\(\sqrt{5x-1}+\sqrt{x}=1\)
=>\(5x-1+x+2\sqrt{x\left(5x-1\right)}=1\)
=>\(6x-1+2\sqrt{5x^2-x}=1\)
=>\(2\sqrt{5x^2-x}=1-6x+1=2-6x\)
=>\(\sqrt{5x^2-x}=1-3x\)
=>\(\begin{cases}1-3x\ge0\\ \left(1-3x\right)^2=5x^2-x\end{cases}\Rightarrow\begin{cases}3x\le1\\ 9x^2-6x+1-5x^2+x=0\end{cases}\)
=>\(\begin{cases}x\le\frac13\\ 4x^2-5x+1=0\end{cases}\Rightarrow\begin{cases}x\le\frac13\\ \left(4x-1\right)\left(x-1\right)=0\end{cases}\Rightarrow x=\frac14\) (nhận)
3: ĐKXĐ: x>=0
Ta có: \(\sqrt{2x+6}-\sqrt{2x}=6\)
=>\(\frac{2x+6-2x}{\sqrt{2x+6}+\sqrt{2x}}=6\)
=>\(\sqrt{2x+6}+\sqrt{2x}=1\)
=>\(2x+2x+6+2\cdot\sqrt{2x\left(2x+6\right)}=1\)
=>\(4\sqrt{x\left(x+3\right)}=1-4x-6=-4x-5\)
=>\(\begin{cases}-4x-5\ge0\\ \left(-4x-5\right)^2=16x\left(x+3\right)\end{cases}\Rightarrow\begin{cases}-4x\ge5\\ 16x^2+40x+25-16x^2-48x=0\end{cases}\)
=>\(\begin{cases}x\le-\frac54\\ -8x+25=0\end{cases}\Rightarrow x\in\) ∅
6: ĐKXĐ: x>=0
Ta có: \(\sqrt{2x+5}-\sqrt{x}=x+5\)
=>\(\frac{2x+5-x}{\sqrt{2x+5}+\sqrt{x}}=x+5\)
=>\(\sqrt{2x+5}+\sqrt{x}=1\)
=>\(2x+5+x+2\sqrt{x\left(2x+5\right)}=1\)
=>\(\sqrt{4x\left(2x+5\right)}=1-3x-5=-3x-4\)
=>\(\begin{cases}-3x-4\ge0\\ 4x\left(2x+5\right)=\left(-3x-4\right)^2\end{cases}\Rightarrow\begin{cases}-3x\ge4\\ 9x^2+24x+16-8x^2-20x=0\end{cases}\)
=>\(\begin{cases}x\le-\frac43\\ x^2-4x+16=0\end{cases}\Rightarrow x\in\) ∅
