Giải:
a) ĐKXĐ: \(x\ne0\)
\(H=1-\dfrac{1}{x^2+4}\left(\left(1-\dfrac{x^2+4}{4x}\right):\left(\dfrac{1}{x}-\dfrac{1}{2}\right)\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\left(\dfrac{4x}{4x}-\dfrac{x^2+4}{4x}\right):\left(\dfrac{4}{4x}-\dfrac{2x}{4x}\right)\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\dfrac{4x-x^2+4}{4x}:\dfrac{4-2x}{4x}\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\dfrac{4x-x^2+4}{4x}.\dfrac{4x}{4-2x}\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{4x-x^2+4}{4-2x}\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{\left(2-x\right)^2}{2\left(2-x\right)}\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{2-x}{2}\)
\(\Leftrightarrow H=1-\dfrac{2-x}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2\left(x^2+4\right)}{2\left(x^2+4\right)}-\dfrac{2-x}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2\left(x^2+4\right)-\left(2-x\right)}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2x^2+8-2+x}{2x^2+8}\)
\(\Leftrightarrow H=\dfrac{2x^2+6+x}{2x^2+8}\)
b) Để \(H=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2x^2+6+x}{2x^2+8}=\dfrac{1}{4}\)
\(\Rightarrow4\left(2x^2+6+x\right)=2x^2+8\)
\(\Leftrightarrow8x^2+24+4x=2x^2+8\)
\(\Leftrightarrow8x^2+24+4x-2x^2-8=0\)
\(\Leftrightarrow6x^2+4x+16=0\)
Đến đây phân tích đa thức thành nhân tử rồi tìm nghiệm.
