Question

Помогите, алгебра 8 класс, спростіть вираз якщоa = 1/3, b= 1/2, вираз вкладений в фото ​

Answer 1

Надеюсь все верно(✿◠‿◠)

Answer 2

Ответ:

12

Объяснение:

$\displaystyle\bf\\\bigg(\frac{a^{-1}}{a^{-1}-b^{-1}} -\frac{a^{-1}-b^{-1}}{a^{-1}} \bigg):\bigg(\frac{b}{a^2} \bigg)^{-1}=\\\\\\=\left(\frac{1}{a\bigg(\dfrac{1}{a}-\dfrac{1}{b}\bigg) } -\frac{a\bigg(\dfrac{1}{a}-\dfrac{1}{b}\bigg) }{1} \right)\cdot\bigg(\frac{b}{a^2} \bigg)=\\\\\\=\left(\dfrac{1}{\dfrac{a(b-a)}{ab} } -\frac{\dfrac{a(b-a)}{ab} }{1} \right)\cdot\frac{b}{a^2} =\\\\\\=\left(\dfrac{1}{\dfrac{b-a}{b} } -\frac{\dfrac{b-a}{b} }{1} \right)\cdot\frac{b}{a^2} =$

$\displaystyle\bf\\=\left(\frac{b}{b-a} -\frac{b-a}{b} \right)\cdot\frac{b}{a^2} =\\\\\\=\frac{b^2-(b-a)^2}{b(b-a)} \cdot\frac{b}{a^2} =\\\\\\=\frac{b^2-b^2+2ab-a^2}{b-a} \cdot\frac{1}{a^2} =\\\\\\=\frac{2ab-a^2}{b-a} \cdot\frac{1}{a^2} =\frac{a(2b-a)}{a^2(b-a)} =\\\\\\=\frac{2b-a}{a(b-a)} ;a=1/3;b=1/2\\\\\\\frac{2\cdot\dfrac{1}{2}-\dfrac{1}{3} }{\dfrac{1}{3}\bigg(\dfrac{1}{2} -\dfrac{1}{3}\bigg ) } =\frac{1-\dfrac{1}{3} }{\dfrac{1}{3}\cdot\dfrac{1}{6} } =\frac{2}{3}\cdot\frac{3\cdot6}{1} =12$