a)

\(9^{8} = \left(\right. 3^{2} \left.\right)^{8} = 3^{16}\)

\(8^{9} = \left(\right. 2^{3} \left.\right)^{9} = 2^{27}\)

ta có:
\(3^{16} = 3. 3^{15} = 3 \left(\right. 3^{3} \left.\right)^{5} = 3.2 7^{5}\)

\(2^{27} = 2^{2} . 2^{25} = 4. \left(\right. 2^{5} \left.\right)^{5} = 4.3 2^{5}\)

⇒ \(3.2 7^{5} < 4.3 2^{5}\) nên \(3^{16} < 2^{27}\)

\(Vậy9^8<8^9\)

\(9^8<8^9\)

\(3^{39}<11^{20}\)

Đề bài:

\(A = \frac{1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3}}{7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5}}\)

Bước 1: Biến đổi các lũy thừa

Ta rút gọn từng số:

• \(1 2^{3} = \left(\right. 3 \cdot 4 \left.\right)^{3} = 3^{3} \cdot 4^{3} = 3^{3} \cdot \left(\right. 2^{2} \left.\right)^{3} = 3^{3} \cdot 2^{6}\)
• \(12 1^{2} = \left(\right. 1 1^{2} \left.\right)^{2} = 1 1^{4}\)
• \(2 2^{4} = \left(\right. 2 \cdot 11 \left.\right)^{4} = 2^{4} \cdot 1 1^{4}\)
• \(3^{3} = 3^{3}\)
• \(7 5^{2} = \left(\right. 3 \cdot 5^{2} \left.\right)^{2} = 3^{2} \cdot 5^{4}\)
• \(1 1^{4} = 1 1^{4}\)
• \(3 0^{2} = \left(\right. 2 \cdot 3 \cdot 5 \left.\right)^{2} = 2^{2} \cdot 3^{2} \cdot 5^{2}\)
• \(1 1^{5} = 1 1^{5}\)

Bước 2: Thay vào biểu thức

Tử số:

\(1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3} = \left(\right. 3^{3} \cdot 2^{6} \cdot 1 1^{4} \cdot 5 \left.\right) - \left(\right. 2^{4} \cdot 1 1^{4} \cdot 3^{3} \left.\right)\)

Rút chung:

\(= 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 2^{2} \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 4 \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 20 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19\)

Mẫu số:

\(7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5} = \left(\right. 3^{2} \cdot 5^{4} \cdot 1 1^{4} \left.\right) - \left(\right. 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 1 1^{5} \left.\right)\)

Rút chung \(3^{2} \cdot 5^{2} \cdot 1 1^{4}\):

\(= 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 5^{2} - 2^{2} \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 4 \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 44 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)\)

Bước 3: Viết lại biểu thức đầy đủ

\(A = \frac{3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19}{3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)}\)

Rút gọn:

• \(3^{3} / 3^{2} = 3\)
• \(1 1^{4}\) triệt tiêu
• \(19 / \left(\right. - 19 \left.\right) = - 1\)

Còn lại:

\(A = \frac{3 \cdot 2^{4}}{5^{2}} \cdot \left(\right. - 1 \left.\right) = \frac{3 \cdot 16}{25} \cdot \left(\right. - 1 \left.\right) = \frac{48}{25} \cdot \left(\right. - 1 \left.\right) = \boxed{- \frac{48}{25}}\)

✅ Kết quả cuối cùng:

\(\boxed{A = - \frac{48}{25}}\)

Ta có: \(12^3\cdot121^2\cdot5-22^4\cdot3^3\)

\(=\left(2^2\cdot3\right)^3\cdot\left(11^2\right)^2\cdot5-11^4\cdot2^4\cdot3^3\)

\(=2^6\cdot3^3\cdot11^4\cdot5-11^4\cdot2^4\cdot3^3=11^4\cdot3^3\cdot2^4\left(2^2\cdot5-1\right)\)

\(=11^4\cdot3^3\cdot2^4\cdot19\)

Ta có: \(75^2\cdot11^4-30^2\cdot11^5\)

\(=\left(3\cdot5^2\right)^2\cdot11^4-\left(2\cdot3\cdot5\right)^2\cdot11^5\)

\(=3^2\cdot5^4\cdot11^4-2^2\cdot3^2\cdot5^2\cdot11^5\)

\(=3^2\cdot5^2\cdot11^4\left(5^2-2^2\cdot11\right)=3^2\cdot5^2\cdot11^4\cdot\left(-19\right)\)

Ta có: \(A=\frac{12^3\cdot121^2\cdot5-22^4\cdot3^3}{75^2\cdot11^4-30^2\cdot11^5}\)

\(=\frac{2^4\cdot3^3\cdot11^4\cdot19}{3^2\cdot5^2\cdot11^4\cdot\left(-19\right)}=\frac{2^4\cdot3}{5^2\cdot\left(-1\right)}=\frac{48}{-25}\)

Bài 1 : 

A = 1 + 2 + 2² + ... + 2¹¹

A = ( 1 + 2 ) + ( 2² + 2³ ) + ... + ( 2¹⁰ + 2¹¹ )

A = 3 + 2²(1+2) + ... + 2¹⁰(1+2)

A = 1.3 + 2².3 + ... + 2¹⁰.3

A = 3.(1+2²+...+2¹⁰) chia hết cho 3

Bài 2 :

2.5² + 3:71⁰ - 54:3³

= 2.25 + 3:1 - 54:27

= 50 + 3 - 2

= 49

Bài 3 :

a) ( 2x - 6 ) . 4⁷ = 4⁹

2x - 6 = 4² = 16

2x = 16

=> x = 8

b) ( 27x + 6 ) : 3 - 11 = 9

( 27x + 6 ) : 3 = 20

27x + 6 = 60

27x = 54

=> x = 2

c) 740 : ( x + 10 ) = 10² - 2.13

740 : ( x + 10 ) = 74

x + 10 = 10

=> x = 0

d) ( 15 - 6x ) . 3⁵ = 3⁶

15 - 6x = 3

6x = 12

=> x = 2

Bài 4 :

Ta có : ab + ba = ( 10a + b ) + ( 10b + a ) = ( 10a + a ) + ( 10b + b ) = 11a + 11a = 11.(a+b) chia hết cho 11 

Bài 1 : 

A = 1 + 2 + 2² + ... + 2¹¹

A = ( 1 + 2 ) + ( 2² + 2³ ) + ... + ( 2¹⁰ + 2¹¹ )

A = 3 + 2²(1+2) + ... + 2¹⁰(1+2)

A = 1.3 + 2².3 + ... + 2¹⁰.3A = 3.(1+2²+...+2¹⁰) chia hết cho 3

Bài 2 :

2.5² + 3:71⁰ - 54:3³

= 2.25 + 3:1 - 54:27

= 50 + 3 - 2= 49

Bài 3 :

a) ( 2x - 6 ) . 4⁷ = 4⁹

2x - 6 = 4² = 16

2x = 16

=> x = 8

b) ( 27x + 6 ) : 3 - 11 = 9

( 27x + 6 ) : 3 = 20

27x + 6 = 60

27x = 54

=> x = 2

c) 740 : ( x + 10 ) = 10² - 2.13

740 : ( x + 10 ) = 74

x + 10 = 10

=> x = 0

d) ( 15 - 6x ) . 3⁵ = 3⁶

15 - 6x = 3

6x = 12

=> x = 2

Bài 4 :

Ta có : ab + ba = ( 10a + b ) + ( 10b + a ) = ( 10a + a ) + ( 10b + b ) = 11a + 11a = 11.(a+b) chia hết cho 11 

b: \(7\cdot2^{13}< 8\cdot2^{13}=2^{16}\)

d: \(3^{99}=\left(3^{33}\right)^3\)

\(11^{21}=\left(11^7\right)^3\)

mà \(3^{33}>11^7\)

nên \(3^{99}>11^{21}\)

\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{\left(3.2^2.2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{\left(3.2^{18}\right)^2}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)

\(=\frac{3^2.2}{11-2}\)

\(=2\)

x=1;y=1

a) Ta có: \(\frac{x}{y}=\frac{3}{11}\Rightarrow\frac{x}{3}=\frac{y}{11}\)

\(\Rightarrow11x=3y\Rightarrow xy=11.3=33.\)

\(\Rightarrow xy=33=1.33=33.1=11.3=3.11 \) (và các số nguyên âm)

Ta lập bảng

b) Tương tự

D=\(\frac{\left(3.4.216\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{3^2.2^4.\left(2^3.3^3\right)^2}{11.2^{13}.2^{22}-2^{36}}=\frac{2^4.3^2.2^6.3^6}{2^{35}\left(11-2\right)}=\frac{2^{10}.3^8}{2^{35}.3^2}\)

D= \(\frac{3^6}{2^{15}}\)

\(D={{3^6} \over {2^{25}}}\) nhé

\(-x-\frac{3}{4}=-\frac{8}{11}=>-x=-\frac{8}{11}+\frac{3}{4}=\frac{1}{44}=>x=-\frac{1}{44}\)

-x-3/4=-8/11

=0.02272727....

11/12-(2/5+x)=2/3

=-0.15