Đáp án: $x=1$
Giải thích các bước giải:
$x\sqrt{{{x}^{2}}-x+1}+2\sqrt{3x+1}={{x}^{2}}+x+3$ (ĐK: $x\ge -\dfrac{1}{3}$)
$\Leftrightarrow 2x\sqrt{{{x}^{2}}-x+1}+4\sqrt{3x+1}=2{{x}^{2}}+2x+6$
$\Leftrightarrow \left[ 2x\sqrt{{{x}^{2}}-x+1}-\left( 3x-1 \right) \right]+\left[ 4\sqrt{3x+1}-\left( 3x+5 \right) \right]=2{{x}^{2}}-4x+2$
$\Leftrightarrow \dfrac{4{{x}^{2}}\left( {{x}^{2}}-x+1 \right)-{{\left( 3x-1 \right)}^{2}}}{2x\sqrt{{{x}^{2}}-x+1}+3x-1}+\dfrac{16\left( 3x+1 \right)-{{\left( 3x+5 \right)}^{2}}}{4\sqrt{3x+1}+3x+5}=2{{x}^{2}}-4x+2$
$\Leftrightarrow \dfrac{{{\left( x-1 \right)}^{2}}\left( 4{{x}^{2}}+4x-1 \right)}{2x\sqrt{{{x}^{2}}-x+1}+3x-1}-\dfrac{9{{\left( x-1 \right)}^{2}}}{4\sqrt{3x+1}+3x+5}=2{{\left( x-1 \right)}^{2}}$
$\Leftrightarrow\left[ \begin{array}{l}\left(x-1\right)^2=0\,\,\,\,\,\left(1\right)\\\dfrac{4{{x}^{2}}+4x-1}{2x\sqrt{{{x}^{2}}-x+1}+3x-1}-\dfrac{9}{4\sqrt{3x+1}+3x+5}=2\,\,\,\,\,\left(2\right)\end{array} \right.$
$\left( 1 \right)\Leftrightarrow x=1$ (nhận)
$\left( 2 \right)\Leftrightarrow \dfrac{4{{x}^{2}}+4x-1}{2x\sqrt{{{x}^{2}}-x+1}+3x-1}=2+\dfrac{9}{4\sqrt{3x+1}+3x+5}$
Ta có $VP$ luôn lớn hơn $2$
Còn $VT$ luôn nhỏ hơn $2$ …..
Nên $\left( 2 \right)$ sẽ vô nghiệm
Vậy tập nghiệm $S=\left\{ 1 \right\}$
