a: ĐKXĐ: x>=2/3 hoặc x=-2/3
Ta có: \(\sqrt{4x^2-9}=2\sqrt{2x+3}\)
=>\(\sqrt{\left(2x-3\right)\left(2x+3\right)}-2\cdot\sqrt{2x+3}=0\)
=>\(\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>\(\left[\begin{array}{l}2x+3=0\\ 2x-3=4\end{array}\right.=>\left[\begin{array}{l}2x=-3\\ 2x=7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac32\left(nhận\right)\\ x=\frac72\left(nhận\right)\end{array}\right.\)
b: ĐKXĐ: x>=3/5
\(\sqrt{25x^2-9}=2\sqrt{5x-3}\)
=>\(\sqrt{5x-3}\cdot\sqrt{5x+3}-2\cdot\sqrt{5x-3}=0\)
=>\(\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{5x-3}=0\\ \sqrt{5x+3}-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}5x-3=0\\ 5x+3=4\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=3\\ 5x=1\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac35\left(nhận\right)\\ x=\frac15\left(loại\right)\end{array}\right.\)
c: ĐKXĐ; x>=3/2
\(\sqrt{4x^2-9}=2\sqrt{2x-3}\)
=>\(\sqrt{\left(2x-3\right)\left(2x+3\right)}-2\cdot\sqrt{2x-3}=0\)
=>\(\sqrt{2x-3}\left(\sqrt{2x+3}-2\right)=0\)
=>\(\left[\begin{array}{l}2x-3=0\\ \sqrt{2x+3}-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-3=0\\ \sqrt{2x+3}=2\end{array}\right.\)
=>\(\left[\begin{array}{l}2x=3\\ 2x+3=4\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3\\ 2x=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32\left(nhận\right)\\ x=\frac12\left(loại\right)\end{array}\right.\)
d: ĐKXĐ: x>=3
\(\sqrt{x-3}-2\cdot\sqrt{x^2-9}=0\)
=>\(\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x-3}=0\\ 1-2\sqrt{x+3}=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x-3=0\\ 2\cdot\sqrt{x+3}=1\end{array}\right.\)
=>\(\left[\begin{array}{l}x-3=0\\ x+3=\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\left(nhận\right)\\ x=\frac14-3=-\frac{11}{4}\left(loại\right)\end{array}\right.\)
e: ĐKXĐ: x>=2
\(\sqrt{x-2}-2\cdot\sqrt{x^2-4}=0\)
=>\(\sqrt{x-2}\left(1-2\sqrt{x+2}\right)=0\)
=>\(\left[\begin{array}{l}x-2=0\\ 1-2\sqrt{x+2}=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ 2\cdot\sqrt{x+2}=1\end{array}\right.\)
=>\(\left[\begin{array}{l}x=2\\ x+2=\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-\frac74\left(loại\right)\end{array}\right.\)
f: ĐKXĐ: x>=2
\(\sqrt{x-2}-3\sqrt{x^2-4}=0\)
=>\(\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x-2}=0\\ 1-3\sqrt{x+2}=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x-2=0\\ 3\sqrt{x+2}=1\end{array}\right.\)
=>\(\left[\begin{array}{l}x-2=0\\ x+2=\frac19\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-\frac{17}{9}\left(loại\right)\end{array}\right.\)
