Question

\(\dfrac{x-2}{x}+\dfrac{x}{x-1}=\dfrac{11}{6}\)

Answer 1

ĐKXĐ: \(x\notin\left\{0;1\right\}\)

Ta có: \(\dfrac{x-2}{x}+\dfrac{x}{x-1}=\dfrac{11}{6}\)

\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x^2}{x\left(x-1\right)}=\dfrac{11}{6}\)

\(\Leftrightarrow\dfrac{12x^2-18x+12}{6x\left(x-1\right)}=\dfrac{11x\left(x-1\right)}{6x\left(x-1\right)}\)

Suy ra: \(12x^2-18x+12=11x^2-11x\)

\(\Leftrightarrow12x^2-18x+12-11x^2+11x=0\)

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

Vậy: S={4;3}