\(PT\left(x\ne\dfrac{7}{2};\dfrac{-7}{2}\right).\\\Leftrightarrow \dfrac{8x+28-35+10x-4-3x}{\left(7-2x\right)\left(7+2x\right)}=0.\\ \Rightarrow15x=11.\)
\(\Leftrightarrow x=\dfrac{11}{15}\left(TM\right).\)
\(\dfrac{4}{7-2x}=\dfrac{5}{2x+7}-\dfrac{4+3x}{4x^2-49}\left(a\right)\)
Ta có : \(4x^2-49=\left(2x+7\right)\left(2x-7\right)\)
\(\RightarrowĐKXĐ:\left\{{}\begin{matrix}7-2x\ne0\\2x+7\ne0\\2x-7\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{7}{2}\\x\ne-\dfrac{7}{2}\\x\ne\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{7}{2}\\x\ne-\dfrac{7}{2}\end{matrix}\right.\)
\(\left(a\right)\Leftrightarrow\dfrac{-4}{2x-7}=\dfrac{5}{2x+7}-\dfrac{4+3x}{4x^2-49}\)
\(\Leftrightarrow\dfrac{-4\left(2x+7\right)}{\left(2x-7\right)\left(2x+7\right)}=\dfrac{5\left(2x-7\right)}{\left(2x-7\right)\left(2x+7\right)}-\dfrac{4+3x}{\left(2x-7\right)\left(2x+7\right)}\left(a_1\right)\)
- Khử mẫu ta được :
\(\left(a_1\right)\Leftrightarrow-4\left(2x+7\right)=5\left(2x-7\right)-\left(4+3x\right)\)
\(\Leftrightarrow-8x-28=10x-35-4-3x\)
\(\Leftrightarrow-8x-10x+3x=28-35-4\)
\(\Leftrightarrow-15x=-11\)
\(\Leftrightarrow x=\dfrac{11}{15}\left(tmđk\right)\)
Vậy : Phương trình có tập nghiệm \(S=\left\{\dfrac{11}{15}\right\}\)
