Question

Salut,am nevoie Urgent de rezolvare!​

Answer 1

Răspuns:

Ai răspuns atașat pe foaie.

Answer 2

Răspuns:

Explicație pas cu pas:

$\bf a=\Big[4^{11}\cdot\big(20\big)^{20}\Big]: \Big [2^{16}\cdot125\cdot\big (2^{5}\big)^{3}\cdot5^{7}\Big]^{2}+2^{3}+\big(7+7^{2}+7^{3}+..+7^{2021}\big)^0$

$\bf a=\Big[\big(2^{2}\big)^{11}\cdot\big(2^{2}\cdot5\big)^{20}\Big]: \Big [2^{16}\cdot 5^{3}\cdot2^{5\cdot3}\cdot5^{7}\Big]^{2}+2^{3}+1$

$\bf a=\Big(2^{2\cdot 11}\cdot2^{2\cdot20}\cdot5^{20}\Big): \Big(2^{16}\cdot 5^{3}\cdot2^{15}\cdot5^{7}\Big)^{2}+2^{3}+1$

$\bf a=\Big(2^{22}\cdot2^{40}\cdot5^{20}\Big): \Big(2^{16+15}\cdot 5^{3+7}\Big)^{2}+8+1$

$\bf a=\Big(2^{22+40}\cdot5^{20}\Big): \Big (2^{31}\cdot 5^{10}\Big)^{2}+8+1$

$\bf a=\Big(2^{62}\cdot5^{20}\Big): \Big(2^{31\cdot2}\cdot 5^{10\cdot2} \Big)+8+1$

$\bf a=\Big(2^{62}\cdot5^{20}\Big): \Big(2^{62}\cdot 5^{20} \Big)+9$

$\bf a= 1+9$

$\bf \red{\underline{~a=10~}}$

_____________________

$\bf b=(41^2-82+1) :8:5-(2^6-2^5+2^2-2^0)$

$\bf b=(1681-82+1) :8:5-(64-32+4-1)$

$\bf b=1600 :8:5-35$

$\bf b=200:5-35$

$\bf b=40-35$

$\purple{\underline{~\bf b=5~}}$

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$\bf c=3^{90}-4\cdot9^{44}+2^{208}:\big(2\cdot4^{103}\big)-5\cdot\big(9^{22}\big)^2$

$\bf c=3^{90}-2^{2} \cdot\big(3^{2}\big)^{44}+2^{208}:\Big[2\cdot\big(2^{2}\big)^{103}\Big]-5\cdot 9^{22\cdot2}$

$\bf c=3^{90}-2^{2} \cdot 3^{2\cdot44}+2^{208}:\Big(2\cdot2^{2\cdot103}\Big)-5\cdot 9^{44}$

$\bf c=3^{90}-2^{2} \cdot 3^{88}+2^{208}:\Big(2\cdot2^{206}\Big)-5\cdot \big(3^{2}\big)^{44}$

$\bf c=3^{90}-2^{2} \cdot 3^{88}+2^{208}:2^{206+1}-5\cdot 3^{2\cdot44}$

$\bf c=3^{90}-4\cdot 3^{88}+2^{208}:2^{207}-5\cdot 3^{88}$

$\bf c=3^{90}-4\cdot 3^{88}+2^{208-207}-5\cdot 3^{88}$

$\bf c=3^{90}-4\cdot 3^{88}+2^{1}-5\cdot 3^{88}$

$\bf c=3^{90}-9\cdot 3^{88}+2$.

$\bf c=3^{90}-3^{2}\cdot 3^{88}+2$

$\bf c=3^{90}-3^{2+88}+2$

$\bf c=3^{90}-3^{90}+2$

$\pink{\underline{~\bf c=2~}}$

$==pav38==$