Question

Va rog dau coroana și toate punctele mele ex 6 a b c e ​

Answer 1

Rezolvarea la a) este sus. Scuze ca nu am făcut și celelalte puncte, dar am aflat ca mai am teme de făcut scuzeeee

Answer 2

Răspuns:

Explicație pas cu pas:

$\bf a) ~40-125:\bigg[5+2^{19}:2^{17}+\Big(2^{2}\Big)^{2}\bigg]=$

$\bf 40-125:\Big(5+2^{19-17}+2^{2\cdot 2}\Big)=$

$\bf 40-125:\Big(5+2^{2}+2^{4}\Big)=$

$\bf 40-125:\Big(5+4+16\Big)=$

$\bf 40-125:25=$

$\bf 40-5=$

$\purple{\boxed{\bf ~35~}}$

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$\bf b) ~\bigg[\Big(5^{4}\Big)^{2}-5^{3}\bigg]:\Big(5^{2}\cdot4\Big)=$

$\bf \Big(5^{4\cdot 2}-5^{3}\Big):\Big(25\cdot4\Big)=$

$\bf \Big(5^{8}-5^{3}\Big):100=$

$\bf 5^{3}\cdot\Big(5^{8-3}-5^{3-3}\Big):100=$

$\bf 5^{3}\cdot\Big(5^{5}-5^{0}\Big):100=$

$\bf 125\cdot\Big(3125-1\Big):100=$

$\bf 125\cdot3124:100=$

$\bf 390500:100=$

$\red{\boxed{~\bf 3905~}}$

___________________

$\bf e) ~\bigg[\Big(17 -4^{2}\Big)^{2012}-\Big(65-4^{3}\Big)^{2011}\bigg]^{6}=$

$\bf \bigg[\Big(17 -16\Big)^{2012}-\Big(65-64\Big)^{2011}\bigg]^{6}=$

$\bf\Big(1^{2012}-1^{2011}\Big)^{6}=$

$\bf\Big(1-1\Big)^{6}=$

$\bf 0^{6}=$

$\blue{\boxed{\bf~ 0~}}$