Question

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

Answer 1

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

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right.\)

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

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

\(\Leftrightarrow-x^2+12x+4=16x+4\)

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

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

Vậy \(S=\left\{0\right\}\)

Answer 2

qui đồng là xong lười làm hả