Question

[4 la puterea 11 ×(20)la puterea 20]:[2la puterea 16×125.(2la puterea a 5) la puterea 3 ×5 la puterea 7]totul la a 2+2 la puterea 3 + 1 ...... va rog dau coroana
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Answer 1

Răspuns:

(4¹¹·20²⁰):[2¹⁶·125·(2⁵)³·5⁷]²+2³+1=

[(2²)¹¹·(2²)²⁰·5²⁰]:[2¹⁶·5³·2¹⁵·5⁷]²+8+1=

(2²²·2⁴⁰·5²⁰):(2³²·5⁶·2³⁰·5¹⁴)+8+1=

(2²²⁺⁴⁰·5²⁰):(2³²⁺³⁰·5⁶⁺¹⁴)+8+1=

(2⁶²·5²⁰):(2⁶²·5²⁰)+8+1=

1+8+1=

10

Answer 2

(4¹¹·20²⁰) : [2¹⁶·125 · (2⁵)³ · 5⁷]² + 2³ + 1 =

$\it (4^{11}\cdot20^{20}):\[[2^{16}\cdot125\cdot(2^5)^3\cdot5^7]^2+2^3+1=\\ \\ =(2^{22}\cdot 2^{40}\cdot5^{20}):(2^{16}\cdot5^3\cdot2^{15}\cdot5^7)^2+9\\ \\ =(2^{42}\cdot5^{20}):(2^{31}\cdot5^{10})^2+1=\dfrac{2^{62}\cdot5^{20}}{2^{62}\cdot5^{20}}+9=1+9=10$