Question

làm tính chia

\(\dfrac{4a^{2^{ }}-9b^2}{a^2b^2}:\dfrac{2ax+3bx}{2ab}\)

\(\dfrac{2x}{25-4b^2}:\dfrac{1}{5+2b}\)

\(\dfrac{\left(2-a\right)^2}{2ab}.\dfrac{b}{\left(2-a\right)}+\dfrac{1}{2}\)

\(\dfrac{2b+2}{2b-b^2}:\dfrac{b+1}{b}+\dfrac{2b+2}{3b-6}\)

Answer 1

\(\dfrac{4a^2-9b^2}{a^2b^2}\div\dfrac{2ax+3bx}{2ab}\)

\(=\dfrac{\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2}\times\dfrac{2ab}{x\left(2a+3b\right)}\)

\(=\dfrac{2ab\left(2a-3b\right)\left(2a+3b\right)}{a^2b^2x\left(2a+3b\right)}=\dfrac{4a-6b}{xab}\)

$\frac{2x}{25-4b^2}:\frac{1}{5+2b}$

\(=\dfrac{2x}{\left(5-2b\right)\left(5+2b\right)}\times\dfrac{5+2b}{1}\)

\(=\dfrac{2x\left(5+2b\right)}{\left(5-2b\right)\left(5+2b\right)}=\dfrac{2x}{5-2b}\)

$\frac{{\left(2-a\right)}^2}{2ab}.\frac{b}{\left(2-a\right)}+\frac{1}{2}$

\(=\dfrac{\left(2-a\right)^2b}{2ab\left(2-a\right)}+\dfrac{1}{2}\)

\(=\dfrac{2b-ab}{2ab}+\dfrac{1}{2}\)

\(=\dfrac{2b-ab}{2ab}+\dfrac{ab}{2ab}=\dfrac{2b}{2ab}=\dfrac{1}{a}\)

$\frac{2b+2}{2b-b^2}:\frac{b+1}{b}+\frac{2b+2}{3b-6}$