Question

Tìm x
\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)

Answer 1

Theo đề ta có:

\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)

\(\Leftrightarrow\dfrac{15x}{\left(x-1\right)\left(x+4\right)}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3\left(x-1\right)}\right)\)

ĐKXĐ : \(x\ne1;x\ne-4\)

\(45x-3\left(x-1\right)\left(x+4\right)=36\left(x-1\right)+12\left(x+4\right)\)

\(\Leftrightarrow45x+3x^2-3x+12=36x-36+12x+48\)

\(\Leftrightarrow-3x^2-6x=0\)

\(\Leftrightarrow-3x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=-2\left(TMĐK\right)\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)