Question

\(\frac{3x}{\left(2x^{2}+x+2\right)}+\frac{7x}{\left(2x^{2}+3x+2\right)}=-8\)

Answer 1

Với \(x=0\) ko phải nghiệm

Với \(x\ne0\) pt trở thành:

\(\dfrac{3}{2x+1+\dfrac{2}{x}}+\dfrac{7}{2x+3+\dfrac{2}{x}}=-8\)

Đặt \(2x+1+\dfrac{2}{x}=t\)

\(\Rightarrow\dfrac{3}{t}+\dfrac{7}{t+2}=-8\)

\(\Rightarrow3\left(t+2\right)+7t=-8t\left(t+2\right)\)

\(\Leftrightarrow8t^2+26t+6=0\Rightarrow\left[{}\begin{matrix}t=-3\\t=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x+1+\dfrac{2}{x}=-3\\2x+1+\dfrac{2}{x}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x^2+4x+2=0\\2x^2+\dfrac{5}{4}x+2=0\end{matrix}\right.\)

\(\Rightarrow x=-1\)