Giải pt: `sqrt{3x^2-12x+13}+sqrt{2x^2-8x+12}=3`

Đáp án: $x=2$

Giải thích các bước giải:

$\sqrt{3{{x}^{2}}-12x+13}+\sqrt{2{{x}^{2}}-8x+12}=3$

$\Leftrightarrow \left( \sqrt{3{{x}^{2}}-12x+13}-1 \right)+\left( \sqrt{2{{x}^{2}}-8x+12}-2 \right)=0$

$\Leftrightarrow \dfrac{3{{x}^{2}}-12x+12}{\sqrt{3{{x}^{2}}-12x+13}+1}+\dfrac{2{{x}^{2}}-8x+8}{\sqrt{2{{x}^{2}}-8x+12}+2}=0$

$\Leftrightarrow \dfrac{3{{\left( x-2 \right)}^{2}}}{\sqrt{3{{x}^{2}}-12x+13}+1}+\dfrac{2{{\left( x-2 \right)}^{2}}}{\sqrt{2{{x}^{2}}-8x+12}+2}=0$

$\Leftrightarrow {{\left( x-2 \right)}^{2}}\left( \dfrac{3}{\sqrt{3{{x}^{2}}-12x+13}+1}+\dfrac{2}{\sqrt{2{{x}^{2}}-8x+12}+2} \right)=0$

$\Leftrightarrow x=2$ (vế sau luôn dương)