Đáp án: $x=\dfrac13$
Giải thích các bước giải:
ĐKXĐ: $x\le -1$ hoặc $x\ge -\dfrac14$
Ta có:
$\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3$
$\to \sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=9x-3$
$\to \dfrac{4x^2+5x+1-\left(4x^2-4x+4\right)}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}=9x-3$
$\to \dfrac{9x-3}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}=9x-3$
$\to \dfrac{9x-3}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}-\left(9x-3\right)=0$
$\to \left(9x-3\right)\left(\dfrac{1}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}-1\right)=0$
$\to 9x-3=0\to x=\dfrac13$
Hoặc $\dfrac{1}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}-1=0$
$\to \dfrac{1}{\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}}=1$
$\to \sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}=1$
$\to \sqrt{4x^2+5x+1}+2\sqrt{x^2-x+1}=1$
$\to \sqrt{4x^2+5x+1}-1=-2\sqrt{x^2-x+1}$
$\to 2\sqrt{4x^2+5x+1}-1=\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}$
$\to 2\sqrt{4x^2+5x+1}-1=9x-3$
$\to 2\sqrt{4x^2+5x+1}=9x-2$
$\to 4\left(4x^2+5x+1\right)=\left(9x-2\right)^2$
$\to 16x^2+20x+4=81x^2-36x+4$
$\to -65x^2+56x=0$
$\to x\left(-65x+56\right)=0$
$\to x=0$ hoặc $x=\dfrac{56}{65}$
Thử lại $\to x=\dfrac13$
