\(\dfrac{1}{32}\) = 2^-5

\(\dfrac{1}{8}\) = 2^-3

0,5 = \(\dfrac{1}{2}\) = 2^-1

0,25 = \(\dfrac{1}{4}\) = 2^-2

\(\dfrac{1}{32}\) = 2^-5

\(\left(-27\right)^5:32^3\)

\(=\left[\left(-3\right)^3\right]^5:\left(2^5\right)^3\)

\(=\left(-3\right)^{15}:2^{15}\)

\(=\left(-3:2\right)^{15}\)

\(=\left(-\dfrac{3}{2}\right)^{15}\)

\(\left(-27\right)^5\):\(32^3\)

=-14348907:32768

=-437,8938904

Bài 1

1) -1/4 + 3/4 : 6

= -1/4 + 1/8

= -1/8

2) (-1/6 - 5/13 + 11/12) - (8/13 - 7/6 - 1/3)

= -1/6 - 5/13 + 11/12 - 8/13 + 7/6 + 1/3

= (-1/6 + 7/6) + (-5/13 - 8/13) + (11/12 + 1/3)

= 1 - 1 + 5/4

= 5/4

3) (-2/3)² . 3/5 + √(4/25) : 9/4 - 1,75

= 4/9 . 3/5 + 2/5 . 4/9 - 7/4

= 4/9 . (3/5 + 2/5) - 7/4

= 4/9 - 7/4

= -47/36

Bài 2

1) -1 2/5 - x = 2/3 - 4/5

-7/5 - x = -2/15

x = -7/5 + 2/15

x = -19/15

2) [√(9/64)+ x] - 1 1/2 = -3/4

3/8 + x - 3/2 = -3/4

x = -3/4 - 3/8 + 3/2

x = 3/8

3) 2.3ˣ⁺² + 4.3ˣ⁺¹ = 10.3⁶

3ˣ⁺¹.(2.3 + 4) = 10.3⁶

3ˣ⁺¹ . 10 = 3⁶ . 10

3ˣ⁺¹ = 3⁶

x + 1 = 6

x = 6 - 1

x = 5

(0,8)⁵ : (0,4)⁶

= (4/5)⁵ : (2/5)⁶

= 1024/3125 : 64/15625

= 16 . 5

= 80

\(\left(0,8\right)^5:\left(0,4\right)^6\\ =\left(\dfrac{4}{5}\right)^5:\left(\dfrac{2}{5}\right)^6\\ =\dfrac{1024}{3125}:\dfrac{64}{15625}\\ =16.5\\ =80\)

\(\left(\dfrac{11}{4}\right)^{12}:\left(-\dfrac{11}{4}\right)^{11}\)

\(=\left(-\dfrac{11}{4}\right)^{12}:\left(-\dfrac{11}{4}\right)^{11}\)

\(=\left(-\dfrac{11}{4}\right)^{12-11}\)

\(=\left(-\dfrac{11}{4}\right)^1\)

\(=-\dfrac{11}{4}\)

(11/4)^12:(-11/4)^11

(3/7)² . (-7)⁴

= (3/7)² . 7⁴

= (3/7)² . 7² . 7²

= 9 . 49

= 441

\(\left(\dfrac{3}{7}\right)^2.\left(-7\right)^4\\ =\left(\dfrac{3}{7}\right)^2.7^4\\ =\left(\dfrac{3}{7}\right)^2.7^2.7^2\\ =9.49\\ =441\)

Vì \(\dfrac{1}{11}>\dfrac{1}{18}>\dfrac{1}{21}>\dfrac{1}{24}>\dfrac{1}{27}>\dfrac{1}{29}\)

\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}>\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}\)\(\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}=\dfrac{1}{11}.7=\dfrac{7}{11}\)

Ta có:

\(\dfrac{7}{11}=\dfrac{7.5}{11.5}=\dfrac{35}{55};\dfrac{4}{5}=\dfrac{4.11}{5.11}=\dfrac{44}{55}\)

\(Vì\) \(\dfrac{44}{55}>\dfrac{35}{55}\)

\(\Rightarrow\dfrac{4}{5}>\dfrac{7}{11}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< \dfrac{4}{5}\left(đpcm\right)\)

Ta thấy :

\(\dfrac{1}{4}+\dfrac{1}{11}< \dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{18}+\dfrac{1}{21}< \dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{24}+\dfrac{1}{27}< \dfrac{1}{24}+\dfrac{1}{24}=\dfrac{1}{12}=\dfrac{1}{3}-\dfrac{1}{4}\)

\(\dfrac{1}{29}< \dfrac{1}{20}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{5}=\dfrac{4}{5}\)

\(\Rightarrow dpcm\)

a) \(a\left(b+1\right)=3\left(a;b\inℤ\right)\)

\(\Rightarrow a;\left(b+1\right)\in U\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-1;-4\right);\left(1;2\right);\left(-3;-2\right);\left(3;0\right)\right\}\)

b) \(2n+7⋮n+1\left(n\inℤ\right)\)

\(\Rightarrow2n+7-2\left(n+1\right)⋮n+1\)

\(\Rightarrow2n+7-2n-2⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow n\in\left\{-2;0;-6;4\right\}\)

c) \(xy+x-y=6\left(x;y\inℤ\right)\)

\(\Rightarrow x\left(y+1\right)-y-1+1=6\)

\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-0;-6\right);\left(2;4\right);\left(-4;-2\right);\left(6;0\right)\right\}\)

Xét \(\Delta ADC\) và \(\Delta BCD\) có:

\(AD=BC\left(gt\right)\)

\(\widehat{ADC}=\widehat{BCD}\left(gt\right)\)

Cạnh \(DC\) là cạnh chung

Vậy \(\Delta ADC=\Delta BCD\left(c.g.c\right)\)

\(\Rightarrow\widehat{C1}=\widehat{D1}\) ( 2 góc tưng ứng )

Trong \(\Delta OCB\) ta có: \(\widehat{C1}=\widehat{D1}\left(CMT\right)\)

\(\Rightarrow\Delta OCB\) cân tại \(O\)

\(\Rightarrow OC=OD\) ( cạnh tương ứng ) \(\left(1\right)\)

\(\Rightarrow AC=BD\) ( tính chất hình thang cân )

\(\Rightarrow AO+OC=BO+OD\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(AO=BO\)

Hình đây nhé !