Đáp án:
$x=-1$
Giải thích các bước giải:
$3(x+1)\sqrt{x^2+x+3}-3x^2-4x-7=0$
$3(x+1)\sqrt{x^2+x+3}-(x+1).(3x-7)=0$
$(x+1).(3\sqrt{x^2+x+3}-(3x-7))=0$
\(\left[ \begin{array}{l}x=-1\\3\sqrt{x^2+x+3}=3x-7\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\\sqrt{x^2+x+3}=x-\dfrac{7}{3}\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\\sqrt{x^2+x+3}=x-\dfrac{7}{3}\\Đk:x\geq \dfrac{7}{3}\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\x^2+x+3=\Big(x-\dfrac{7}{3}\Big)^2\\Đk:x\geq \dfrac{7}{3}\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\x^2+x+3=x^2-\dfrac{14}{3}x+\dfrac{49}{9}\\Đk:x\geq \dfrac{7}{3}\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\\dfrac{17}{3}x=\dfrac{22}{9}\\Đk:x\geq \dfrac{7}{3}\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\17.9x=22.3\\Đk:x\geq \dfrac{7}{3}\end{array} \right.\)
\(\left[ \begin{array}{l}x=-1\\x=\dfrac{51}{22}(loại)\\Đk:x\geq \dfrac{7}{3}\end{array} \right.\)
Vậy $x=-1$
