a)

\(175\cdot19+38\cdot175+43\cdot175\\ =175\cdot19+175\cdot38+175\cdot43\\ =175\cdot\left(19+38+43\right)\\ =175\cdot100\\ =17500\)

b)

\(125\cdot75+125\cdot13-80\cdot125\\ =125\cdot75+125\cdot13-125\cdot80\\ =125\cdot\left(75+13-80\right)\\ =125\cdot10\\ =125\cdot8\\ =1000\)

a, 175. 19 + 38. 175 + 43. 175

= 175. 19 + 175. 38 + 175. 43

= 175.(19 + 38 + 43)

= 175. 100

= 17500 

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

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

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: \(x^3=125\)

=>\(x^3=5^3\)

=>x=5

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

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

=>x+1=2

=>x=2-1=1

c: \(\left(x-5\right)^3=27\)

=>\(\left(x-5\right)^3=3^3\)

=>x-5=3

=>x=3+5=8

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

=>\(4^{x}=4^5:16=4^5:4^2=4^3\)

=>x=3

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

=>\(5^{x-1}=1000:8=125=5^3\)

=>x-1=3

=>x=3+1=4

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

=>\(2^{x}+2^{x}\cdot8=72\)

=>\(2^{x}\cdot9=72\)

=>\(2^{x}=\frac{72}{9}=8=2^3\)

=>x=3

g: \(\left(3x+1\right)^3=343\)

=>\(\left(3x+1\right)^3=7^3\)

=>3x+1=7

=>3x=6

=>x=2

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

=>\(3^{x}+3^{x}\cdot9=270\)

=>\(10\cdot3^{x}=270\)

=>\(3^{x}=\frac{270}{10}=27=3^3\)

=>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

=>x=1

Câu 6:

1 giờ=3600 giây

Số tế bào hồng cầu được tạo ra sau mỗi giờ là:

\(25\cdot10^5\cdot3600=25\cdot36\cdot10^7=900\cdot10^7=9\cdot10^9\) =9 tỉ (tế bào)

câu 5:

a. \(16^{16}=\left(2^4\right)^{16}=2^{64}\)

\(64^{11}=\left(2^6\right)^{11}=2^{66}\)

vì \(2^{66}>2^{64}\) nên \(64^{11}>16^{16}\)

b. \(625^5=\left(5^4\right)^5=5^{20}\)

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

\(5^{20}<5^{21}\Rightarrow625^5<125^7\)

c. \(3^{36}=\left(3^3\right)^{12}=27^{12}\)

\(5^{24}=\left(5^2\right)^{12}=25^{12}\)

\(27^{12}>25^{12}\Rightarrow3^{36}>5^{24}\)

Ta có: \(\frac{A}{10^{10}}=\frac{10^{20}-6}{10^{20}-6\cdot10^{10}}=\frac{10^{20}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}\)

\(\frac{B}{10^{10}}=\frac{10^{21}-6}{10^{21}-6\cdot10^{10}}=\frac{10^{21}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)

Ta có: \(10^{20}<10^{21}\)

=>\(10^{20}-6\cdot10^{10}<10^{21}-6\cdot10^{10}\)

=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)

=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}+1>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}+1\)

=>\(\frac{A}{10^{10}}>\frac{B}{10^{10}}\)

=>A>B

b lớn hon a

ko tin thì thôi tui lớp 6

67:

a: \(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac83\)

=>\(\frac{-6x}{12}+\frac{8x}{12}+\frac{3\left(x+1\right)}{12}+\frac{2\left(2x+1\right)}{12}=\frac{32}{12}\)

=>-6x+8x+3(x+1)+2(2x+1)=32

=>2x+3x+3+4x+2=32

=>9x=32-5=27

=>x=3

b: \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

=>\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{12}{26}=\frac{6}{13}\)

=>\(\frac{6}{2x+1}=\frac{6}{13}\)

=>2x+1=13

=>2x=12

=>x=6

Bài 68:

a: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};...;\frac{1}{100}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{50}{50}=1\) (1)

Ta có: \(\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+..+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}\)

=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac{50}{100}=\frac12\) (2)

Từ (1),(2) suy ra \(\frac12<\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<1\)

b: Ta có: \(\frac{1}{21}<\frac{1}{20};\frac{1}{22}<\frac{1}{20};\ldots;\frac{1}{30}<\frac{1}{20}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}<\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}=\frac{10}{20}=\frac12\) (3)

Ta có: \(\frac{1}{31}<\frac{1}{30};\frac{1}{32}<\frac{1}{30};\ldots;\frac{1}{40}<\frac{1}{30}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}<\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (4)

Từ (3),(4) suy ra \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}<\frac12+\frac13=\frac56\left(5\right)\)

Ta có: \(\frac{1}{21}>\frac{1}{30};\frac{1}{22}>\frac{1}{30};\ldots;\frac{1}{30}=\frac{1}{30}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (6)

Ta có: \(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};\ldots;\frac{1}{40}=\frac{1}{40}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{10}{40}=\frac14\) (7)

Từ (6),(7) suy ra \(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}>\frac13+\frac14=\frac{7}{12}\) (8)

Từ (5),(8) suy ra \(\frac{7}{12}<\frac{1}{21}+\ldots+\frac{1}{40}<\frac56\)

Lời giải:
a.

$=\frac{3}{5}-\frac{7}{4}=\frac{12-35}{20}=\frac{-23}{20}$

b.

$=-(2+\frac{5}{8})=-\frac{21}{8}$

c.

$=-(\frac{1}{8}+\frac{5}{9})=-\frac{9+8.5}{8.9}=\frac{-49}{72}$
d.

$=\frac{6}{13}-\frac{14}{39}=\frac{18}{39}-\frac{14}{39}=\frac{4}{39}$

e.

$=\frac{-3}{4}+\frac{5}{7}=\frac{5}{7}-\frac{3}{4}$

$=\frac{20-21}{7.4}=\frac{-1}{28}$

Bài 5

1) x ∈ Ư(18) = {1; 2; 3; 6; 9; 18}

x ∈ B(4) = {0; 4; 8; 12; 16; 20; ...}

Vậy không tìm được x thỏa mãn đề bài

2) x ∈ Ư(20) = {1; 2; 4; 5; 10; 20}

x ∈ B(2) = {0; 2; 4; 6; 8; 10; 12; 14; 16; 18; 20; ...}

⇒ x ∈ {2; 4; 10; 20}

3) x ∈ B(12) = {0; 12; 24; 36; 48; ...; 96; 108; ...}

Mà 30 ≤ x ≤ 100

⇒ x ∈ {36; 48; ...; 96}

4) x ∈ Ư(150) = {1; 2; 3; 5; 6; 10; 15; 25; 30; 50; 75; 150}

Mà x ≤ 50

⇒ x ∈ {1; 2; 3; 5; 6; 10; 15; 25; 30; 50}

5) 70 ⋮ x và 168 ⋮ x

⇒ x ∈ ƯC(70; 168)

Ta có:

70 = 2.5.7

168 = 2³.3.7

⇒ ƯCLN(70; 168) = 2.7 = 14

⇒ x ∈ ƯC(70; 168) = Ư(14) = {1; 2; 7; 14}

Mà x > 10

⇒ x = 14

6) Ta có:

(1995 + 2005 + x) ⋮ 5

1995 ⋮ 5

2005 ⋮ 5

⇒ x ⋮ 5

⇒ x ∈ B(5) = {0; 5; 10; 15; 20; 25; 30; 35; 40; ...}

Mà 23 < x ≤ 35

⇒ x ∈ {25; 30; 35}

Bài 6

1) Do 17x2y chia hết cho 2 và 5 nên y = 0

⇒ Số đã cho có dạng: 17x20

Để 17x20 chia hết cho 3 thì (1 + 7 + x + 2 + 0) ⋮ 3

⇒ (10 + x) ⋮  3

⇒ x ∈ {2; 5; 8}

Vậy x ∈ {2; 5; 8}; y = 0

2) Do 234xy chia hết cho 2 và 5 nên y = 0

⇒ Số đã cho có dạng: 234x0

Để 234x0 chia hết cho 9 thì (2 + 3 + 4 + x + 0) ⋮ 9

⇒ (9 + x) ⋮ 9

⇒ x ∈ {0; 9}

Vậy x ∈ {0; 9}; y = 0

3) Do 4x6y chia hết cho 2 và 5 nên y = 0

Mà x - y = 4

⇒ x = 4 + y

⇒ x = 4

Vậy x = 4; y = 0

4) Do 57x2y chia hết cho 5 nhưng không chia hết cho 2 nên y = 5

⇒ Số đã cho có dạng 57x25

Để 57x25 chia hết cho 9 thì (5 + 7 + x + 2 + 5) ⋮ 9

⇒ (19 + x) ⋮ 9

⇒ x = 8

Vậy x = 8; y = 5