\(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=9x-3\)(ĐKXĐ : \(x\le-1\)hoặc \(x\ge-\frac{1}{4}\))
\(\Leftrightarrow\left(\sqrt{4x^2+5x+1}-2\sqrt{7}x\right)-\left(\sqrt{4x^2-4x+4}-2\sqrt{7}x\right)-\left(9x-3\right)=0\)
\(\Leftrightarrow\frac{\left(4x^2+5x+1\right)-28x^2}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-\frac{\left(4x^2-4x+4\right)-28x^2}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-3\left(3x-1\right)=0\)
\(\Leftrightarrow\frac{-24x^2+5x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}+\frac{24x^2+4x-4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-3\left(3x-1\right)=0\)
\(\Leftrightarrow\frac{-\left(3x-1\right)\left(8x+1\right)}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}+\frac{4\left(3x-1\right)\left(2x+1\right)}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-3\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-1=0\\\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3=0\end{array}\right.\)
Với 3x - 1 = 0 => x = \(\frac{1}{3}\) (TMĐK)
Với \(\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3=0\) , Từ điều kiện \(\left[\begin{array}{nghiempt}x\le-1\\x\ge-\frac{1}{4}\end{array}\right.\) ta luôn có : \(\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3>0\)
Vậy phương trình có nghiệm : \(x=\frac{1}{3}\)
