ĐKXĐ x>=-\(\frac{-3}{2}\)
Bình phương
4x2-9=4(2x+3)
4x2-9-8x-12=0
4x2-8x-20=0
\(x=1-\sqrt{6}\)hoặc\(x=1+\sqrt{6}\)
\(\sqrt{4x^2-9}=2\sqrt{2x+3}\)
\(\Rightarrow\sqrt{\left(2x-3\right)\left(2x+3\right)}=2\sqrt{2x+3}\)
\(\Rightarrow\frac{\sqrt{\left(2x-3\right)}.\sqrt{2x+3}}{\sqrt{2x+3}}=2\)
\(\Rightarrow\sqrt{\left(2x-3\right)}=2\)
\(\Rightarrow\sqrt{\left(2x-3\right)}=\sqrt{4}\)
\(\Rightarrow2x-3=4\)\(\Rightarrow2x=7\)\(\Rightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)thì thỏa mãn điều kiện trên
Thiếu: ĐKXĐ: \(x\ge\frac{-3}{2}\)