9 x 5 = 45

63 : 7 = 9

8 x 8 = 64

5 x 7 = 35

8 x 7 =56

3 x 8 = 24

40 : 5 = 8

5 x 5 = 25

7 x 5 = 35

7 x 8 =56

6 x 4 = 24

45 : 9 = 5

7 x 7 = 49

35 : 5 = 7

56 : 8 = 7

2 x 8 = 16

81 : 9 = 9

9 x 9 = 81

35 : 7 = 5

56 : 7 = 8

ay bạn ơi dễ vậy mà

1. 

$(5^{1986}-5^{1985}):5^{1985}=5^{1985}(5-1):5^{1985}=5-1=4$

2.

\((7^{846}+7^{847}):7^{846}=7^{846}(1+7):7^{846}=1+7=8\)

3.

\((9^{2018}-3^{4036}):6^{2006}=[(3^2)^{2018}-3^{4036}]:6^{2006}\)

$=(3^{4036}-3^{4036}):6^{2006}=0:6^{2006}=0$

4.

$(7^{80}.8^{70}-56^{70}):56^{70}$

$=[7^{10}(7.8)^{70}-56^{70}]:56^{70}$

$=[7^{10}.56^{70}-56^{70}]:56^{70}$

$=56^{70}(7^{10}-1):56^{70}=7^{10}-1$

5.

$4^{4016}:(4^{4017}-4^{4016})=4^{4016}:[4^{4016}(4-1)]$

$=4^{4016}:4^{4016}:3=1:3=\frac{1}{3}$

6.

$(12^{206}.2^{207}-24^{206}):24^{206}$

$=(12^{206}.2^{206}.2-24^{206}):24^{206}$

$=[(12.2)^{206}.2-24^{206}]:24^{206}$

$=(24^{206}.2-24^{206}):24^{206}$

$=24^{206}(2-1):24^{206}=2-1=1$

7.

$(5^2-24)^{8946}+4^{30}:2^{60}=1^{8946}+(2^2)^{30}:2^{60}$

$=1+2^{60}:2^{60}=1+1=2$

8.

$(37.8^{1007}-7.2^{3021}):8^{1007}=[37.8^{1007}-7.(2^3)^{1007}]:8^{1007}$

$=[37.8^{1007}-7.8^{1007}]:8^{1007}$

$=8^{1007}(37-7):8^{1007}=37-7=30$

= 1192 nhé

 hok tốt

:D

9 × 5 = 45

6 × 4 = 24

63 : 7 = 9

40 : 5 = 8

8 × 8 = 64

7 × 5 = 35

56 : 8 = 7

42 : 6 = 7

\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

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