Đặt \(\sqrt{x-4}=a\left(a\ge0\right)\Rightarrow x-4=a^2\Rightarrow x=a^2+4\) . Khi đó , ta có :
\(3\left(a^2+4\right)+2\left(a+6\right)=12\sqrt{a^2+4}\)
\(\Leftrightarrow3a^2+2a+24=12\sqrt{a^2+4}\)
\(\Leftrightarrow\left(3a^2+2a+24\right)^2=144\left(a^2+4\right)\)
\(\Leftrightarrow9a^4+4a^2+576+12a^3+144a^2+96a=144a^2+576\)
\(\Leftrightarrow9a^4+12a^3+4a^2+96a=0\)
\(\Leftrightarrow a\left(9a^3+12a^2+4a+96\right)=0\)
\(\Leftrightarrow a\left[3a^2\left(3a+8\right)-4a\left(3a+8\right)+12\left(3a+8\right)\right]=0\)
\(\Leftrightarrow a\left(3a+8\right)\left(3a^2-4a+12\right)=0\)
Do : \(3a^2-4a+12=3\left(a^2-\frac{4}{3}a+\frac{4}{9}\right)+\frac{32}{3}>0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\3a+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=-\frac{8}{3}\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x-4}=0\Leftrightarrow x=4\)
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