Question

Tính :

\(\frac{2}{2x+3}\) + \(\frac{5}{2x-3}\) - \(\frac{2x-33}{9-4x^2}\)

Mik cần gấp giúp với !

Answer 1

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)

= \(\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

= \(\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

= \(\frac{4x-6+10x-15+2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

= \(\frac{16x-54}{\left(2x-3\right)\left(2x+3\right)}\)

Answer 2

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)\(=\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

\(=\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{2x-33}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\frac{4x-6+10x+15+2x-33}{\left(2x+3\right)\left(2x-3\right)}=\frac{16x-24}{\left(2x+3\right)\left(2x-3\right)}=\frac{8\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}=\frac{8}{2x+3}\)