Question

\(\frac{4}{4x^2-8x+7}+\frac{3}{4x^2-10x+7}=\frac{1}{x}\)

Answer 1

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\frac{4x}{4x^2-8x+7}+\frac{3x}{4x^2-10x+7}=1\)

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

Đặt \(4x-8+\frac{7}{x}=a\) phương trình trở thành:

\(\frac{4}{a}+\frac{3}{a-2}=1\) \(\Leftrightarrow a\left(a-2\right)=4\left(a-2\right)+3a\)

\(\Leftrightarrow a^2-9a+8=0\Rightarrow\left[{}\begin{matrix}a=1\\a=8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-8+\frac{7}{x}=1\\4x-8+\frac{7}{x}=8\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\left(vn\right)\\4x^2-16x+7=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{7}{2}\end{matrix}\right.\)