Question

\(\dfrac{4}{-25x^2+20x-3}=\dfrac{3}{5x-1}-\dfrac{2x}{5x-3}\)

Answer 1

Đk: \(x\ne\dfrac{3}{5};x\ne\dfrac{1}{5}\)

Pt \(\Leftrightarrow\dfrac{4}{\left(5x-3\right)\left(1-5x\right)}=\dfrac{-3\left(5x-3\right)}{\left(1-5x\right)\left(5x-3\right)}-\dfrac{2x\left(1-5x\right)}{\left(1-5x\right)\left(5x-3\right)}\)

\(\Rightarrow4=-3\left(5x-3\right)-2x\left(1-5x\right)\)

\(\Leftrightarrow-10x^2+17x-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17+\sqrt{89}}{20}\left(tmpt\right)\\x=\dfrac{17-\sqrt{89}}{20}\left(ktmpt\right)\end{matrix}\right.\)

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