Question

\(\sqrt{2x^2-x+3}\) - \(\sqrt{21x-17}\)=x-x²

Answer 1

ĐKXĐ: \(x\ge\dfrac{17}{21}\)

\(\Leftrightarrow x^2-3x+2+\left(\sqrt{2x^2-x+3}-\left(x+1\right)\right)+\left(3x-1-\sqrt{21x-17}\right)=0\)

\(\Leftrightarrow x^2-3x+2+\dfrac{x^2-3x+2}{\sqrt{2x^2-x+3}+x+1}+\dfrac{9\left(x^2-3x+2\right)}{3x-1+\sqrt{21x-17}}=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{1}{\sqrt{2x^2-x+3}+x+1}+\dfrac{9}{3x-1+\sqrt{21x-17}}\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)