\(\Leftrightarrow2\sqrt{x^2-x+5}=x^2-x-3\)
Đặt \(x^2-x-3=a\)
Pt sẽ là \(2\sqrt{a+8}=a\)
\(\Leftrightarrow\sqrt{4a+32}=a\)
\(\Leftrightarrow\left\{{}\begin{matrix}a>=0\\a^2=4a+32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>=0\\\left(a-8\right)\left(a+4\right)=0\end{matrix}\right.\Leftrightarrow a=8\)
=>\(x^2-x-3=8\)
=>\(x^2-x-11=0\)
hay \(x\in\left\{\dfrac{1+3\sqrt{5}}{2};\dfrac{1-3\sqrt{5}}{2}\right\}\)
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