Question

$10*{18^{2}:324+2*[(2^{2}*3 )^{15}:(2^{29}*3^{15} )+1^{24}]}$
Fundă De la 10* incepe paranteza { si se gata la ]}

Answer 1

Răspuns: 70

Explicație pas cu pas:

$\bf \bf 10 \cdot\Big \{18^{2}:324 + 2\cdot\big[(2^{2}\cdot3)^{15}:(2^{29}\cdot3^{15})+1^{24}\big]\Big \} =$

$\bf 10 \cdot \Big\{324:324 + 2\cdot\big[(2^{2\cdot15} \cdot3^{15}):(2^{29}\cdot3^{15})+1\big]\Big \} =$

$\bf 10 \cdot \Big\{1 + 2\cdot\big[(2^{30} \cdot3^{15}):(2^{29}\cdot3^{15})+1\big]\Big \} =$

$\bf 10 \cdot\big[1+ 2\cdot(2^{30 - 29} \cdot3^{15 - 15}+1)\big]=$

$\bf 10 \cdot\big[1+ 2\cdot(2^{1} \cdot3^{0}+1)\big]=$

$\bf 10 \cdot\big[1+ 2\cdot(2 \cdot 1+1)\big]=$

$\bf 10 \cdot\big[1+ 2\cdot(2+1)\big]=$

$\bf 10 \cdot\big(1+ 2\cdot 3 \big)=$

$\bf 10 \cdot\big(1+ 6 \big)=$

$\bf 10 \cdot 7 =$

$\pink{ \underline{ \: \bf 70 \: }}$

$==pav38==$