Question

Thực hiện phép tính \(\left(\frac{1}{3+2x}+\frac{1}{3-2x}\right)\div\frac{1}{3+2x}\)      giúp mình vs,minh cần gấp

Answer 1

DK \(3+2x\ne0;3-2x\ne0\Leftrightarrow x\ne\left\{+-\frac{3}{2}\right\}\)

\(\left(\frac{3-2x+\left(3+2x\right)}{\left(3+2x\right)\left(3-2x\right)}\right).\frac{\left(3+2x\right)}{1}\)\(=\left(\frac{6}{\left(3+2x\right)\left(3-2x\right)}\right).\left(3+2x\right)\)\(=\frac{6}{\left(3-2x\right)}\)

Answer 2

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

\(=\frac{6}{3-2x}\)