Question

Giải phương trình : 

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

b) \(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)

Answer 1

ĐK \(y\ne\left\{-\frac{1}{3};\frac{1}{3};3\right\}\)

a. Ta có \(\frac{1}{3y^2-10y+3}=\frac{6y}{9y^2-1}+\frac{2}{1-3y}\)

\(\frac{\Leftrightarrow1}{\left(y-3\right)\left(3y-1\right)}=\frac{6y}{\left(3y+1\right)\left(3y-1\right)}-\frac{2}{3y-1}\)

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

\(\Leftrightarrow3y+1=-2y+6\Leftrightarrow5y=5\Rightarrow y=1\)

Vậy \(y=1\)

b. Pt \(\Leftrightarrow x-\frac{\frac{x-3}{4}}{2}=3-\frac{\frac{x-3}{6}}{2}\Leftrightarrow x-\frac{x-3}{8}=3-\frac{x-3}{12}\)

\(\Leftrightarrow\left(x-3\right)-\frac{x-3}{8}-\frac{x-3}{12}=0\Leftrightarrow\frac{19}{24}\left(x-3\right)=0\Leftrightarrow x=3\)

Vậy \(x=3\)