Question

113. Giải phương trình: \(\dfrac{1}{3y^2-10y}=\dfrac{6y}{9y^2-1}+\dfrac{2}{1-3y}\)

Answer 1

\(\dfrac{1}{3y^2-10y}=\dfrac{6y}{9y^2-1}+\dfrac{2}{1-3y}\)

\(\Leftrightarrow\dfrac{1}{3y^2-10y}=\dfrac{6y-2\left(3y+1\right)}{\left(3y-1\right)\left(3y+1\right)}\)

\(\Leftrightarrow\dfrac{1}{3y^2-10y}=\dfrac{-2}{9y^2-1}\)

\(\Leftrightarrow9y^2-1=-6y^2+20y\)

\(\Leftrightarrow15y^2-20y-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{10+\sqrt{115}}{15}\\y=\dfrac{10-\sqrt{115}}{15}\end{matrix}\right.\)