Question

giải pt

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

Answer 1

Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne-2\end{matrix}\right.\)

\(\frac{12}{x^2-4}+\frac{1}{2x-x^2}=\frac{4+x}{x\left(x+2\right)}\)\(\Leftrightarrow \dfrac{12x}{x(x-2)(x+2)}+\dfrac{(x+2)}{x(2-x)(x+2}-\dfrac{4x(x-2)}{x(x+2)(x-2)}=0\)

\(\Leftrightarrow \dfrac{12x-(x+2)-4x(x-2)}{x(x-2)(x+2)}=0\)

\(\Leftrightarrow -4x^2+11x=0\)\(\Leftrightarrow x\left(-4x+11\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\-4x+11=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{11}{4}\end{matrix}\right.\)

KL:........................