Question

Efectuati:[1 1/2+1,1(6)+1,(3)]×[(1/3)^2+(3/2)^5:(3/2)^3-13/36]​

Answer 1

$\bf \Big[1\dfrac{1}{2} +1,1(6)+1,(3)\Big]\cdot\Big[\Big(\dfrac{1}{3}\Big)^{2}+\Big(\dfrac{3}{2}\Big)^{5}:\Big(\dfrac{3}{2}\Big)^{3}-\dfrac{13}{36}\Big]=$

$\bf \Big[\dfrac{3}{2} +\dfrac{105}{90}+\dfrac{12}{9}\Big]\cdot\Big[\Big(\dfrac{1}{3}\Big)^{2}+\Big(\dfrac{3}{2}\Big)^{5}:\Big(\dfrac{3}{2}\Big)^{3}-\dfrac{13}{36}\Big]=$

$\bf \dfrac{3\cdot 90+105\cdot2+12\cdot20}{180} \cdot\Big(\dfrac{1}{9}+\Big(\dfrac{3}{2}\Big)^{5-3}-\dfrac{13}{36}\Big)=$

$\bf \dfrac{504}{180} \cdot\Big(\dfrac{1}{9}+\Big(\dfrac{3}{2}\Big)^{2}-\dfrac{13}{36}\Big)=$

$\bf \dfrac{\not504}{\not180} \cdot\Big(\dfrac{1}{9}+\dfrac{9}{4}-\dfrac{13}{36}\Big)=$

$\bf 4 \cdot\Big(\dfrac{1\cdot4}{36}+\dfrac{9\cdot9}{36}-\dfrac{13}{36}\Big)=$

$\bf 4 \cdot\dfrac{4+81-13}{36}=$

$\bf 4 \cdot\dfrac{72}{36}=$

$\bf 4 \cdot\dfrac{\not72}{\not36}=$

$\bf 4 \cdot 2=$

$\boxed{\bf 8}$