a)
\(ĐKXĐ:\left\{{}\begin{matrix}2x+3\ne0\\2x+1\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\dfrac{2}{2x+3}+\dfrac{3}{2x+1}-\dfrac{6x+5}{\left(2x+3\right)\left(2x+1\right)}\)
\(=\dfrac{2\left(2x+1\right)}{\left(2x+3\right)\left(2x+1\right)}+\dfrac{3\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}-\dfrac{6x+5}{\left(2x+3\right)\left(2x+1\right)}\)
\(=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}\\ =\dfrac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\\ =\dfrac{2}{2x+1}\)
3)
với C=-1 ta có
\(-1=\dfrac{2}{2x+1}\\ =>-1\cdot\left(2x+1\right)=2\\ =>2x+1=-2\\ =>2x=-3\\ =>x=-\dfrac{3}{2}\left(loai\right)\)
vậy x∈{∅}
giúp mk vs ạ ai nhanh mk tick nha 
