Question

b \(\dfrac{4}{2x-3}+\dfrac{4x}{4x^2-1}=\dfrac{7}{2x-1}\)

Answer 1

sai đề

Answer 2

Sửa đề: \(\dfrac{4}{2x+1}+\dfrac{4x}{4x^2-1}=\dfrac{7}{2x-1}\)

ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{4}{2x+1}+\dfrac{4x}{4x^2-1}=\dfrac{7}{2x-1}\)

\(\Leftrightarrow\dfrac{4\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{4x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{7\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)

Suy ra: \(8x-4+4x=14x+7\)

\(\Leftrightarrow12x-4-14x-7=0\)

\(\Leftrightarrow-2x-11=0\)

\(\Leftrightarrow-2x=11\)

hay \(x=-\dfrac{11}{2}\)(thỏa ĐK)

Vậy: \(S=\left\{-\dfrac{11}{2}\right\}\)