2: ĐKXĐ: \(x\ge\frac23\)
Ta có: \(\sqrt{x+3}=3\cdot\sqrt{3x-2}-\sqrt{2x-1}\)
=>\(\sqrt{x+3}+\sqrt{2x-1}=3\cdot\sqrt{3x-2}=\sqrt{9\left(3x-2\right)}=\sqrt{27x-18}\)
=>\(\left(\sqrt{x+3}+\sqrt{2x-1}\right)^2=27x-18\)
=>\(x+3+2x-1+2\cdot\sqrt{\left(x+3\right)\left(2x-1\right)}=27x-18\)
=>\(2\sqrt{\left(x+3\right)\left(2x-1\right)}=27x-18-3x-2=24x-20\)
=>\(\sqrt{\left(x+3\right)\left(2x-1\right)}=12x-10\)
=>\(\begin{cases}12x-10\ge0\\ \left(12x-10\right)^2=\left(x+3\right)\left(2x-1\right)\end{cases}\Rightarrow\begin{cases}x\ge\frac56\\ 144x^2-240x+100=2x^2+5x-3\end{cases}\)
=>\(\begin{cases}x\ge\frac56\\ 142x^2-245x+103=0\end{cases}\Rightarrow\begin{cases}x\ge\frac56\\ 142x^2-142x-103x+103=0\end{cases}\)
=>\(\begin{cases}x\ge\frac56\\ \left(x-1\right)\left(142x-103\right)=0\end{cases}\Rightarrow\begin{cases}x\ge\frac56\\ x\in\left\lbrace1;\frac{103}{142}\right\rbrace\end{cases}\)
=>x=1
3: ĐKXĐ: x>=-1/5
Ta có: \(\sqrt{5x+1}-\sqrt{2x+3}=\sqrt{14x+7}\)
=>\(\left(\sqrt{5x+1}-\sqrt{2x+3}\right)^2=\left(\sqrt{14x+7}\right)^2\)
=>\(5x+1+2x+3-2\cdot\sqrt{\left(5x+1\right)\left(2x+3\right)}=14x+7\)
=>\(7x+4-2\cdot\sqrt{\left(5x+1\right)\left(2x+3\right)}=14x+7\)
=>\(2\sqrt{\left(5x+1\right)\left(2x+3\right)}=7x+4-14x-7=-7x-3\)
=>\(\sqrt{4\left(5x+1\right)\left(2x+3\right)}=-7x-3\)
=>\(\begin{cases}4\left(5x+1\right)\left(2x+3\right)=\left(-7x-3\right)^2\\ -7x-3>=0\end{cases}\Rightarrow\begin{cases}49x^2+42x+9=4\left(10x^2+15x+2x+3\right)\\ -7x\ge3\end{cases}\)
=>\(\begin{cases}49x^2+42x+9=40x^2+68x+12\\ x\le-\frac37\end{cases}\Rightarrow\begin{cases}9x^2-26x-3=0\\ x\le-\frac37\end{cases}\)
=>\(\begin{cases}\left(x-3\right)\left(9x+1\right)=0\\ x\le-\frac37\end{cases}\Rightarrow\begin{cases}x\in\left\lbrace3;-\frac19\right\rbrace\\ x\le-\frac37\end{cases}\)
=>x∈∅
