\(\dfrac{3y-1}{3y+1}+\dfrac{y-3}{y+3}=2\)
\(\Leftrightarrow\dfrac{\left(3y-1\right)\left(y+3\right)}{\left(3y+1\right)\left(y+3\right)}+\dfrac{\left(y-3\right)\left(3y+1\right)}{\left(y+3\right)\left(3y+1\right)}=\dfrac{2\left(3y+1\right)\left(y+3\right)}{\left(3y+1\right)\left(y+3\right)}\)
\(\Rightarrow\left(3y+1\right)\left(y+3\right)+\left(y-3\right)\left(3y+1\right)=2\left(3y+1\right)\left(y+3\right)\)
\(\Leftrightarrow\left(3y+1\right)\left(y+3\right)+\left(y-3\right)\left(3y+1\right)-2\left(3y+1\right)\left(y+3\right)=0\)
\(\Leftrightarrow\left(3y+1\right)\left(y+3\right)+\left(y-3\right)\left(3y+1\right)-\left(3y+1\right)\left(2y+6\right)=0\)
\(\Leftrightarrow\left(3y+1\right)\left(y+3+y-3-2y-6\right)=0\)
\(\Leftrightarrow-6\left(3y+1\right)=0\)
\(\Leftrightarrow3y+1=0\)
\(\Leftrightarrow3y=-1\)
\(\Leftrightarrow y=\dfrac{-1}{3}\)
Vậy \(S=\left\{\dfrac{-1}{3}\right\}\)
\(\dfrac{3y-1}{3y+1}+\dfrac{y-3}{y+3}=2\) ĐKXĐ: y\(\ne\dfrac{-1}{3},y\ne-3\)
\(\Rightarrow\left(3y-1\right)\left(y+3\right)+\left(y-3\right)\left(3y+1\right)=2\left(3y+1\right)\left(y+3\right)\)
\(\Leftrightarrow3y^2+9y-y-3+3y^2+y-9y-3=6y^2+18y+2y+6\)
\(\Leftrightarrow3y^2+3y^2-6y^2-18y-2y=6+3+3\)
\(\Leftrightarrow-20y=12\)
\(\Leftrightarrow y=\dfrac{-3}{5}\) (TĐK)
\(\Rightarrow S=\left\{\dfrac{-3}{5}\right\}\)
Thiếu đkxđ :))
đkxđ : y \(\ne\dfrac{-1}{3}\)
y \(\ne-3\)
Vậy S = \(\varnothing\)
