Răspuns:
Explicație pas cu pas:
[tex]\bf a) ~40-125:\bigg[5+2^{19}:2^{17}+\Big(2^{2}\Big)^{2}\bigg]=[/tex]
[tex]\bf 40-125:\Big(5+2^{19-17}+2^{2\cdot 2}\Big)=[/tex]
[tex]\bf 40-125:\Big(5+2^{2}+2^{4}\Big)=[/tex]
[tex]\bf 40-125:\Big(5+4+16\Big)=[/tex]
[tex]\bf 40-125:25=[/tex]
[tex]\bf 40-5=[/tex]
[tex]\purple{\boxed{\bf ~35~}}[/tex]
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[tex]\bf b) ~\bigg[\Big(5^{4}\Big)^{2}-5^{3}\bigg]:\Big(5^{2}\cdot4\Big)=[/tex]
[tex]\bf \Big(5^{4\cdot 2}-5^{3}\Big):\Big(25\cdot4\Big)=[/tex]
[tex]\bf \Big(5^{8}-5^{3}\Big):100=[/tex]
[tex]\bf 5^{3}\cdot\Big(5^{8-3}-5^{3-3}\Big):100=[/tex]
[tex]\bf 5^{3}\cdot\Big(5^{5}-5^{0}\Big):100=[/tex]
[tex]\bf 125\cdot\Big(3125-1\Big):100=[/tex]
[tex]\bf 125\cdot3124:100=[/tex]
[tex]\bf 390500:100=[/tex]
[tex]\red{\boxed{~\bf 3905~}}[/tex]
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[tex]\bf e) ~\bigg[\Big(17 -4^{2}\Big)^{2012}-\Big(65-4^{3}\Big)^{2011}\bigg]^{6}=[/tex]
[tex]\bf \bigg[\Big(17 -16\Big)^{2012}-\Big(65-64\Big)^{2011}\bigg]^{6}=[/tex]
[tex]\bf\Big(1^{2012}-1^{2011}\Big)^{6}=[/tex]
[tex]\bf\Big(1-1\Big)^{6}=[/tex]
[tex]\bf 0^{6}=[/tex]
[tex]\blue{\boxed{\bf~ 0~}}[/tex]
