Question

f (12-3)^2*(6+3)^18-9^20
g (4^2:3^2)^2*4^18:256^5
Dau coroana
$< 3$
​

Answer 1

Răspuns:

b.

7¹⁺²³⁻²⁴=7⁰=1

c.

11⁵⁰-11¹⁺⁴⁹=11⁵⁰-11⁵⁰=0

d.

2¹⁷⁺¹⁸-2³⁴⁺¹=2³⁵-2³⁵=0

e.

(29-14)²·15²⁰-15²²=15²⁺²⁰-15²²=0

f.

(12-3)²·(6+3)¹⁸-9²⁰=9²·9¹⁸-9²⁰=9²⁺¹⁸-9²⁰=0

g.

(4²:3²)²*4¹⁸:256⁵=[(2²)²:2²]²·(2²)¹⁸:(2⁸)⁵ =[2²ˣ²:2²]²·2²ˣ¹⁸:2⁸ˣ⁵ =(2⁴:2²)²·2³⁶:2⁴⁰=

(2⁴⁻²)²·2³⁶:2⁴⁰ =2²ˣ²·2³⁶:2⁴⁰=2⁴⁺³⁶⁻⁴⁰ =1

Answer 2

Explicație pas cu pas:

b)

$7 \cdot {7}^{23} : {7}^{24} = {7}^{1 + 23} : {7}^{24} = {7}^{24} : {7}^{24} = {7}^{24 - 24} = {7}^{0} = \bf 1$

c)

${11}^{50} - 11 \cdot {11}^{49} = {11}^{50} - {11}^{1 + 49} = {11}^{50} - {11}^{50} = \bf 0$

d)

${2}^{17} \cdot {2}^{18} - {2}^{34} \cdot 2 = {2}^{17 + 18} - {2}^{34 + 1} = {2}^{35} - {2}^{35} = \bf 0$

e)

$(29 - 2 \cdot 7)^{2} \cdot {15}^{20} - {15}^{22} = (29 - 14)^{2} \cdot {15}^{20} - {15}^{22} = (15)^{2} \cdot {15}^{20} - {15}^{22} = (15)^{2 + 20} - {15}^{22} = (15)^{22} - {15}^{22} = \bf 0$

f)

$(12-3)^2 \cdot (6+3)^{18} - 9^{20} = {9}^2 \cdot {9}^{18} - {9}^{20} = {9}^{2 + 18} - {9}^{20} = {9}^{20} - {9}^{20} = \bf 0$

g)

$(4^2:3^2)^2 \cdot 4^{18}:256^5 = {4}^{4} : {3}^{4} \cdot 4^{18} : ( {4}^{4} )^{5} = {4}^{4 + 18} : {3}^{4} : {4}^{20} = {4}^{22} : {3}^{4} : {4}^{20} = {4}^{22 - 20} : {3}^{4} = {4}^{2} : {3}^{4} = \bf \frac{16}{81}$