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Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{4x^2+5x+1}=b\ge0\end{matrix}\right.\) ta được:
\(b-2a=4a^2-b^2\)
\(\Leftrightarrow\left(2a-b\right)\left(2a+b\right)+2a-b=0\)
\(\Leftrightarrow\left(2a-b\right)\left(2a+b+1\right)=0\)
\(\Leftrightarrow2a-b=0\Rightarrow b=2a\)
\(\Leftrightarrow b^2=4a^2\Leftrightarrow4x^2+5x+1=4\left(x^2-x+1\right)\)
\(\Leftrightarrow9x=3\Rightarrow x=\frac{1}{3}\)
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