Question

3/x^2-3x+2+2/x^2-4x+3=1/x^2-5x+6

Answer 1

ĐKXĐ: \(x\notin\left\{1;2;3\right\}\)

Ta có: \(\dfrac{3}{x^2-3x+2}+\dfrac{2}{x^2-4x+3}=\dfrac{1}{x^2-5x+6}\)

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(x-1\right)\left(x-3\right)}=\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{x-1}-\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-3}=0\)

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{x-2-x+1}{\left(x-1\right)\left(x-2\right)}=0\)

Suy ra: \(3-1=0\)(Vô lý)

Vậy: \(S=\varnothing\)