Question

.
Calculaţi a) [(4 supra 5 plus 5 supra 6 ori 3 supra 25) :9 pe 2 - 1 pe 15] ori 3 intregi 3 supra 4 + 1 pe 2

Answer 1

Răspuns: $\color{red}\LARGE\boxed{\bf 1}$

Explicație pas cu pas:

$\large \bf \bigg[\bigg(\dfrac{4}{5} +\dfrac{5}{6} \cdot \dfrac{3}{25}\bigg):\dfrac{9}{2}-\dfrac{1}{15}\bigg]\cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf \bigg[\bigg(\dfrac{4}{5} +\dfrac{\not5}{\not 6} \cdot \dfrac{\not3}{\not25}\bigg)\cdot \dfrac{2}{9}-\dfrac{1}{15}\bigg]\cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf \bigg[\bigg(\dfrac{4}{5} +\dfrac{1}{2} \cdot \dfrac{1}{5}\bigg)\cdot \dfrac{2}{9}-\dfrac{1}{15}\bigg]\cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf \bigg[\bigg(\dfrac{8}{10} +\dfrac{1}{10}\bigg)\cdot \dfrac{2}{9}-\dfrac{1}{15}\bigg]\cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf \bigg(\dfrac{9}{10} \cdot \dfrac{2}{9}-\dfrac{1}{15}\bigg)\cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf \bigg(\dfrac{\not9}{\not10} \cdot \dfrac{\not2}{\not9}-\dfrac{1}{15}\bigg)\cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf \bigg(\dfrac{1}{5} -\dfrac{1}{15}\bigg)\cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf\dfrac{2}{15} \cdot \dfrac{15}{4} +\dfrac{1}{2} =$

$\large \bf\dfrac{\not2}{\not15} \cdot \dfrac{\not15}{\not4} +\dfrac{1}{2} =$

$\large \bf\dfrac{1}{2} +\dfrac{1}{2} =$

$\color{red}\LARGE\boxed{\bf 1}$

==pav38==