Question

giải pt \(\left(x+4\right)\sqrt{3x^2+1}=x^2+x+4\)

Answer 1

Bạn coi lại căn thức 

Answer 2

\(\Leftrightarrow\left(x+4\right)\left(\sqrt{3x^2+1}-1\right)=x^2\)

\(\Leftrightarrow\dfrac{\left(x+4\right).3x^2}{\sqrt{3x^2+1}+1}=x^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{3\left(x+4\right)}{\sqrt{3x^2+1}+1}=1\left(1\right)\end{matrix}\right.\)

Xét (1):

\(\Leftrightarrow3x+12=\sqrt{3x^2+1}+1\)

\(\Leftrightarrow3x+11=\sqrt{3x^2+1}\) (\(x\ge-\dfrac{11}{3}\))

\(\Leftrightarrow9x^2+66x+121=3x^2+1\)

\(\Leftrightarrow6x^2+66x+120=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{2}\\x=\dfrac{-11-\sqrt{41}}{2}\left(loại\right)\end{matrix}\right.\)