Bài 8:

a: \(5^3=125;3^5=243\)

mà 125<243

nên \(5^3<3^5\)

b: \(7\cdot2^{13}<8\cdot2^{13}=2^3\cdot2^{13}=2^{16}\)

c: \(27^5=\left(3^3\right)^5=3^{3\cdot5}=3^{15}\)

\(243^3=\left(3^5\right)^3=3^{5\cdot3}=3^{15}\)

Do đó: \(27^5=243^5\)

d: \(625^5=\left(5^4\right)^5=5^{4\cdot5}=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{3\cdot7}=5^{21}\)

mà 20<21

nên \(625^5<125^7\)

Bài 9:

a: \(3^{x}\cdot5=135\)

=>\(3^{x}=\frac{135}{5}=27=3^3\)

=>x=3(nhận)

b: \(\left(x-3\right)^3=\left(x-3\right)^2\)

=>\(\left(x-3\right)^3-\left(x-3\right)^2=0\)

=>\(\left(x-3\right)^2\cdot\left\lbrack\left(x-3\right)-1\right\rbrack=0\)

=>\(\left(x-3\right)^2\cdot\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\left(nhận\right)\\ x=4\left(nhận\right)\end{array}\right.\)

c: \(\left(2x-1\right)^4=81\)

=>\(\left[\begin{array}{l}2x-1=3\\ 2x-1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=4\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-1\left(loại\right)\end{array}\right.\)

d: \(\left(5x+1\right)^2=3^2\cdot5+76\)

=>\(\left(5x+1\right)^2=9\cdot5+76=45+76=121\)

=>\(\left[\begin{array}{l}5x+1=11\\ 5x+1=-11\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=10\\ 5x=-12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-\frac{12}{5}\left(loại\right)\end{array}\right.\)

e: \(5+2^{x-3}=29-\left\lbrack4^2-\left(3^2-1\right)\right\rbrack\)

=>\(2^{x-3}+5=29-\left\lbrack16-9+1\right\rbrack\)

=>\(2^{x-3}+5=29-8=21\)

=>\(2^{x-3}=16=2^4\)

=>x-3=4

=>x=4+3=7(nhận)

f: \(3+2^{x-1}=24-\left\lbrack4^2-\left(2^2-1\right)\right\rbrack\)

=>\(2^{x-1}+3=24-\left\lbrack16-4+1\right\rbrack=24-13=11\)

=>\(2^{x-1}=11-3=8=2^3\)

=>x-1=3

=>x=4(nhận)

Bài 6:

a: \(5\cdot5\cdot5\cdot5\cdot5\cdot5=5^6\)

b: \(27\cdot14\cdot7\cdot2=27\cdot14\cdot14=3^3\cdot14^2\)

c: \(x\cdot x\cdot x\cdot y=x^3\cdot y\)

d: \(5^3\cdot5^4=5^{3+4}=5^7\)

e: \(7^8:7^2=7^{8-2}=7^6\)

f: \(42^7:6^7\cdot49=7^7\cdot49=7^7\cdot7^2=7^{7+2}=7^9\)

Câu 8:

a:Sửa đề: \(4+4^2+\cdots+4^{2025}\)

Ta có: \(4+4^2+\cdots+4^{2025}\)

\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+\cdots+\left(4^{2023}+4^{2024}+4^{2025}\right)\)

\(=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+\cdots+4^{2023}\left(1+4+4^2\right)\)

\(=21\left(4+4^4+\cdots+4^{2023}\right)\) ⋮21

b: \(5+5^2+5^3+5^4+\cdots+5^{2024}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{2023}+5^{2024}\right)\)

\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+\cdots+5^{2022}\left(5+5^2\right)\)

\(=30\left(1+5^2+\cdots+5^{2022}\right)\) ⋮30

Câu 7:

a: \(A=2+2^2+2^3+\cdots+2^{99}\)

=>\(2A=2^2+2^3+\cdots+2^{100}\)

=>\(2A-A=2^2+2^3+\cdots+2^{100}-2-2^2-\cdots-2^{99}\)

=>\(A=2^{100}-2\)

b: \(B=1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)

=>\(7B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}\)

=>\(7B+B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}+1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)

=>\(8B=-7^{50}+1\)

=>\(B=\frac{-7^{50}+1}{8}\)

câu 4:

a) \(\)x³ = 125

x³ = 5³

⇒ x = 5

b. \(11^{x+1}=121\)

\(11^{x+1}=11^2\)

⇒ x + 1 = 2

⇒ x = 2 - 1 = 1

c. (x - 5)³ = 27

(x - 5)³ = 3³

⇒ x - 5 = 3

x = 3 + 5 = 8

d. \(4^5:4^{x}=16\)

\(4^{5-x}=4^2\)

⇒ 5 - x = 2

x = 5 - 2 = 3

e. \(5^{x-1}\cdot8=1000\)

\(5^{x-1}=1000:8\)

\(5^{x-1}=125\)

\(5^{x-1}=5^3\)

⇒ x - 1 = 3

x = 3 + 1 = 4

f. \(2^{x}+2^{x+3}=72\)

\(2^{x}\cdot\left(1+2^3\right)=72\)

\(2^{x}=72:9\)

\(2^{x}=8\)

\(2^{x}=2^3\)

⇒ x = 3

g. (3x + 1)³ = 343

(3x + 1)³ = 7³

⇒ 3x + 1 = 7

3x = 7 - 1

3x = 6

x = 6 : 3 = 2

h. \(3^{x}+3^{x+2}=270\)

\(3^{x}\cdot\left(1+3^2\right)=270\)

\(3^{x}=270:10\)

\(3^{x}=27\)

\(3^{x}=3^3\)

⇒ x = 3

i. \(25^{2x+4}=125^{x+3}\)

\(\left(5^2\right)^{2x+4}=\left(5^3\right)^{x+3}\)

\(5^{4x+8}=5^{3x+9}\)

=>4x + 8 = 3x + 9

4x - 3x = 9 - 8

x = 1

c: \(\left(x-1\right)^3=\left(-9\right)^3\)

=>x-1=-9

=>x=-9+1=-8

f: \(3x-2^3=7+\left(-9\right)\)

=>3x-8=7-9=-2

=>3x=-2+8=6

=>x=2

a) diện tích △ ADG là:

20 x 9 : 2 = 90 (cm2)

diện tích △ ABE là:

14 x 8 : 2 = 56 (cm2)

diện tích hình chữ nhật ABCD là:

20 x 14 = 280 (cm2)

diện tích tứ giác AECG là:

280 - 56 - 90 = 134 (cm2)

b) tỉ số diện tích △ ABE và diện tích △ ADG là:

\(\frac{56}{90}=\frac{28}{45}\)

Ta có: \(10A=\frac{10^{21}-60}{10^{21}-6}=\frac{10^{21}-6-54}{10^{21}-6}=1-\frac{54}{10^{21}-6}\)

\(10B=\frac{10^{22}-60}{10^{22}-6}=\frac{10^{22}-6-54}{10^{22}-6}=1-\frac{54}{10^{22}-6}\)

Ta có: \(10^{21}-6<10^{22}-6\)

=>\(\frac{54}{10^{21}-6}>\frac{54}{10^{22}-6}\)

=>\(-\frac{54}{10^{21}-6}<-\frac{54}{10^{22}-6}\)

=>\(-\frac{54}{10^{21}-6}+1<-\frac{54}{10^{22}-6}+1\)

=>10A<10B

=>A<B

Bài 3:

4; 45 + 5\(x\) = 10\(^3\): 10

45 + 5\(x\) = 100

5\(x\) = 100 - 45

5\(x\) = 55

\(x\) = 55 : 5

\(x\) = 11

Vậy \(x=11\)

5; 4\(x\) - 20 = 2\(^5\) : 2\(^2\)

4\(x\) - 20 = 2\(^3\)

4\(x\) = 8 + 20

4\(x\) = 28

\(x\) = 28 : 4

\(x=7\)

Vậy \(x=7\)

Bài 4:

1; 82 - (25 + 4\(x^{}\)) = 17

25 + 4\(x\) \(^{}\) = 82 - 17

4\(x^{}\) = 65 - 25

4\(x^{}\) = 40

\(x=40:4\)

\(x\) = 10

Vậy \(x=10\)

2; 71 - (24 + 3\(x\)) = 24

24 + 3\(x\) = 71 - 24

24 + 3\(x\) = 47

3\(x\) = 47 - 24

3\(x\) = 23

\(x\) = 23 : 3

Vậy \(x=\frac{23}{3}\)

3; 145 - (125 + \(x\)) = 12

125 + \(x\) = 145 - 12

125 + \(x\) = 133

\(x\) = 133 - 125

\(x\) = 8

Vậy \(x=8\)

Ta có các quy luật sau:

\(\left(1+3\right)-2=2\)

\(\left(2+2\right)-3=1\)

\(\left(5+5\right)-6=4\)

Vậy dòng cuối là: 

\(\left(5+9\right)-5=9\)

Số điền vào là 9

(Quy luật: lấy 2 số phía dưới cộng với nhau rồi trừ cho số phía trên sẽ ra được số ở giữa)

( 1 + 3 ) − 2 = 2

( 2 + 2 ) − 3 = 1

( 5 + 5 ) − 6 = 4

Ta có dòng cuối là:

( 5 + 9 ) − 5 = 9

=>Số cần tìm là 9