Đáp án:
$x = 2$
Giải thích các bước giải:
$\sqrt{3x^2 - 12x + 21} +\sqrt{5x^2 - 20x + 24} = -2x^2 + 8x - 3$
$\Leftrightarrow 2x^2 - 8x + 8 + (\sqrt{3x^2 - 12x + 21} - 3) + (\sqrt{5x^2 - 20x + 24} - 2) = 0$
$\Leftrightarrow 2(x^2 - 4x + 4) + \dfrac{(\sqrt{3x^2 - 12x + 21} - 3)(\sqrt{3x^2 - 12x + 21} + 3)}{\sqrt{3x^2 - 12x + 21} + 3} +\dfrac{(\sqrt{5x^2 - 20x + 24} - 2)(\sqrt{5x^2 - 20x + 24} +2)}{\sqrt{5x^2 - 20x + 24} + 2} = 0$
$\Leftrightarrow 2(x^2 - 4x + 4) +\dfrac{3x^2 - 12x + 12}{\sqrt{3x^2 - 12x + 21} + 3} +\dfrac{5x^2 - 20x + 20}{\sqrt{5x^2 - 20x + 24} + 2} = 0$
$\Leftrightarrow (x^2 - 4x + 4)\left(2 + \dfrac{3}{\sqrt{3x^2 - 12x + 21} + 3}+ \dfrac{5}{\sqrt{5x^2 - 20x + 24} + 2}\right) = 0$
$\Leftrightarrow (x-2)^2\cdot\left(2 + \dfrac{3}{\sqrt{3x^2 - 12x + 21} + 3}+ \dfrac{5}{\sqrt{5x^2 - 20x + 24} + 2}\right) = 0$
$\Leftrightarrow \left[\begin{array}{l}x = 2\\2 + \dfrac{3}{\sqrt{3x^2 - 12x + 21} + 3}+ \dfrac{5}{\sqrt{5x^2 - 20x + 24} + 2} =0\quad (*)\end{array}\right.$
Ta có:
$2 + \dfrac{3}{\sqrt{3x^2 - 12x + 21} + 3}+ \dfrac{5}{\sqrt{5x^2 - 20x + 24} + 2} > 0\quad \forall x$
Do đó $(*)$ vô nghiệm
Vậy phương trình có nghiệm duy nhất $x = 2$
