Question

Giải phương trình: \(\dfrac{4}{4x^2-8x+7}+\dfrac{3}{4x^2-10x+7}=\dfrac{1}{x}\)

Answer 1

ĐKXĐ: \(x\ne0\)

Phương trình tương đương:

\(\dfrac{4}{4x-8+\dfrac{7}{x}}+\dfrac{3}{4x-10+\dfrac{7}{x}}=1\)

Đặt \(4x-10+\dfrac{7}{x}=t\)

\(\Rightarrow\dfrac{4}{t+2}+\dfrac{3}{t}=1\)

\(\Rightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)

\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x-10+\dfrac{7}{x}=-1\\4x-10+\dfrac{7}{x}=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x^2-9x+7=0\left(vn\right)\\4x^2-16x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)