Biểu thức \(\dfrac{y-1}{y-2}-\dfrac{y+3}{y-4}\) bằng \(\dfrac{-2}{\left(y-2\right)\left(y-4\right)}\) tức:
\(\dfrac{y-1}{y-2}-\dfrac{y+3}{y-4}=\dfrac{-2}{\left(y-2\right)\left(y-4\right)}\) (ĐK: \(x\ne2;x\ne4\))
\(\Leftrightarrow\dfrac{\left(y-1\right)\left(y-4\right)}{\left(y-2\right)\left(y-4\right)}-\dfrac{\left(y+3\right)\left(y-2\right)}{\left(y-2\right)\left(y-4\right)}=\dfrac{-2}{\left(y-2\right)\left(y-4\right)}\)
\(\Rightarrow\left(y-1\right)\left(y-4\right)-\left(y+3\right)\left(y-2\right)=-2\)
\(\Leftrightarrow y^2-4y-y+4-\left(y^2-2y+3y-6\right)=-2\)
\(\Leftrightarrow y^2-5y+4-y^2-y+6=-2\)
\(\Leftrightarrow\left(y^2-y^2\right)+\left(-5y-y\right)+\left(4+6\right)=-2\)
\(\Leftrightarrow-6y+10=-2\)
\(\Leftrightarrow-6y=-2-10\)
\(\Leftrightarrow-6y=-12\)
\(\Leftrightarrow6y=12\)
\(\Leftrightarrow y=\dfrac{12}{6}\)
\(\Leftrightarrow y=2\left(ktmdk\right)\)
Vậy phương trình vô ngiệm
ĐKXĐ: y<>2; y<>4
\(\dfrac{y-1}{y-2}-\dfrac{y+3}{y-4}=\dfrac{-2}{\left(y-2\right)\left(y-4\right)}\)
=>(y-1)(y-4)-(y+3)(y-2)=-2
=>y^2-5y+4-y^2-y+6=-2
=>-6y=-2-4-6=-12
=>y=2(loại)
