Question

Giải phương trình:

\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{3}{x^2-4x+3}=0\)

Answer 1

\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{3}{x^2-4x+3}=0\\ \Leftrightarrow\frac{1}{\left(x-1\right)\cdot\left(x-2\right)}+\frac{1}{\left(x-2\right)\cdot\left(x-3\right)}-\frac{3}{\left(x-1\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow\frac{x-3}{\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)}+\frac{x-1}{\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)}-\frac{3x-6}{\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow\frac{x-3+x-1-3x+6}{\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow\frac{2-x}{\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow2-x=0\\ \Rightarrow x=2\)