Question

Mn giúp em với ạ

Answer 1

\(a,=\dfrac{4\left(11x+10\right)-3\left(15x+13\right)}{12\left(x-1\right)}=\dfrac{44x+40-45x-39}{12\left(x-1\right)}=\dfrac{1-x}{12\left(x-1\right)}=\dfrac{-1}{12}\\ b,=\dfrac{3a+5+4}{\left(a+3\right)\left(3a+5\right)}=\dfrac{3\left(a+3\right)}{\left(a+3\right)\left(3a+5\right)}=\dfrac{3}{3a+5}\)

Answer 2

a: \(=\dfrac{11x+10}{3\left(x-1\right)}-\dfrac{15x+13}{4\left(x-1\right)}\)

\(=\dfrac{44x+40-45x-39}{12\left(x-1\right)}\)

\(=\dfrac{-x+1}{12\left(x-1\right)}=\dfrac{-1}{12}\)