ta có: \(2x^2+\sqrt{x^2-5x-6}=10x+15\)(đk:\(x\ge6\))
\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)(*)
\(đặt\) \(\sqrt{x^2-5x-6}=a\left(a\ge0\right)\)
=> a2=x2-5x-6
khi đó:
(*) <=>\(2a^2+a-3\) =0
\(\Leftrightarrow\left(2a+3\right)\left(a-1\right)=0\)
\(\Leftrightarrow a-1=0\)(vì 2a+3>0 do a\(\ge0\))
\(\Leftrightarrow a=1\)
\(\Leftrightarrow\sqrt{x^2-5x-6}=1\) \(\Leftrightarrow x^2-5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{53}}{2}\left(tm\right)\\x=\frac{5-\sqrt{53}}{2}\left(loại\right)\end{matrix}\right.\)
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