Question

\(\dfrac{2x-1}{x-4}-\dfrac{x+4}{x+3}=\dfrac{10x+9}{\left(x-4\right)\left(x+3\right)}\)

Answer 1

ĐKXĐ: \(x\ne 4;x\ne -3\)

\(pt\Leftrightarrow\dfrac{\left(2x-1\right)\left(x+3\right)}{\left(x-4\right)\left(x+3\right)}-\dfrac{\left(x+4\right)\left(x-4\right)}{\left(x+3\right)\left(x-4\right)}=\dfrac{10x+9}{\left(x-4\right)\left(x+3\right)}\)

\(\Rightarrow\left(2x-1\right)\left(x+3\right)-\left(x+4\right)\left(x-4\right)=10x+9\)

\(\Leftrightarrow2x^2+5x-3-x^2+16=10x+9\)

\(\Leftrightarrow x^2-5x+4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\left(\text{loại}\right)\end{matrix}\right.\)