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 

2/

Xét phân số \(\dfrac{2n-3}{n+1}=\dfrac{2n+2-5}{n+1}=\dfrac{2n+2}{n+1}-\dfrac{5}{n+1}=\dfrac{2\left(n+1\right)}{n+1}-\dfrac{5}{n+1}=2-\dfrac{5}{n+1}\)

\(n\in Z\Rightarrow2n-3\inƯ\left(5\right)=\left\{-1;-5;1;5\right\}\)

Ta có bảng:

Vậy \(n\in\left\{-1;1;2;4\right\}\)

1/

(x + 1) + (x + 3) + (x + 5) + ... + (x + 999) = 500

<=> (x + x + x + ... + x) + (1 + 3 + 5 + ... + 999) = 500

Xét tổng A = 1 + 3 + 5 + ... + 999

Số số hạng của A là: (999 - 1) : 2 + 1 = 500 

Tổng A là: (999 + 1) x 500 : 2 = 250 000

Do A có 500 số hạng nên có 500 ẩn x.

Vậy ta có: 500x + 250 000 = 500

=> 500x = -249 500

=> x = 499

Vậy x = 499

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 \(\frac{7}{12}<\frac{1}{21}+\ldots+\frac{1}{40}<\frac56\)

Giúp mình với mng ơi!!

1: \(1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\)

\(=1026-82:41\)

=1026-2

=1024

\(2^{11}:\left\lbrace1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\right\rbrace\)

\(=2^{11}:2^{10}=2\)

2: \(250:\left\lbrace1500:\left\lbrack4\cdot5^3-2^3\cdot25\right\rbrack\right\rbrace\)

\(=250:\left\lbrace1500:\left\lbrack4\cdot125-8\cdot25\right\rbrack\right\rbrace\)

\(=250:\left\lbrace1500:\left\lbrack500-200\right\rbrack\right\rbrace=250:\frac{1500}{3}=250:500=0,5\)

3: \(12+3\cdot\left\lbrace90:\left\lbrack39-\left(2^3-5\right)^2\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:\left\lbrack39-\left(8-5\right)^2\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:\left\lbrack39-3^2\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:\left\lbrack39-9\right\rbrack\right\rbrace\)

\(=12+3\cdot\left\lbrace90:30\right\rbrace=12+3\cdot3=21\)

4: \(24:\left\lbrace390:\left\lbrack500-\left(5^3+49\cdot5\right)\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:\left\lbrack500-\left(125+245\right)\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:\left\lbrack500-125-245\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:\left\lbrack375-245\right\rbrack\right\rbrace\)

\(=24:\left\lbrace390:130\right\rbrace=\frac{24}{3}=8\)

5: \(117:\left\lbrace\left\lbrack79-3\cdot\left(3^3-17\right)\right\rbrack:7+2\right\rbrace\)

\(=117:\left\lbrace\left\lbrack79-3\cdot\left(27-17\right)\right\rbrack:7+2\right\rbrace\)

\(=117:\left\lbrace\left\lbrack79-3\cdot10\right\rbrack:7+2\right\rbrace\)

\(=117:\left\lbrace49:7+2\right\rbrace=\frac{117}{9}=13\)

6: \(514-4\cdot\left\lbrace\left\lbrack40+8\left(6-3\right)^2\right\rbrack-12\right\rbrace\)

\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot3^2\right\rbrack-12\right\rbrace\)

\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot9\right\rbrack-12\right\rbrace\)

\(=514-4\cdot\left\lbrace112-12\right\rbrace\)

\(=514-4\cdot100=514-400=114\)

7: \(25\cdot\left\lbrace32:\left\lbrack\left(12-4\right)+4\cdot\left(16:2^3\right)\right\rbrack\right\rbrace\)

\(=25\cdot\left\lbrace32:\left\lbrack8+4\cdot2\right\rbrack\right\rbrace\)

\(=25\cdot\left\lbrace32:16\right\rbrace=25\cdot2=50\)

8: \(30:\left\lbrace175:\left\lbrack355-\left(135+37\cdot5\right)\right\rbrack\right\rbrace\)

\(=30:\left\lbrace175:\left\lbrack355-\left(135+185\right)\right\rbrack\right\rbrace\)

\(=30:\left\lbrace175:\left\lbrack355-320\right\rbrack\right\rbrace=30:\left\lbrace175:35\right\rbrace=\frac{30}{5}=6\)

9: \(32:\left\lbrace160:\left\lbrack300-\left(175+21\cdot5\right)\right\rbrack\right\rbrace\)

\(=32:\left\lbrace160:\left\lbrack300-\left(175+105\right)\right\rbrack\right\rbrace\)

\(=32:\left\lbrace160:\left\lbrack300-280\right\rbrack\right\rbrace\)

\(=32:\left\lbrace160:20\right\rbrace=\frac{32}{8}=4\)

10: \(750:\left\lbrace130-\left\lbrack\left(5\cdot14-65\right)^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-\left\lbrack\left(70-65\right)^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-\left\lbrack5^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-128\right\rbrace=750:2=375\)

Ta có: \(F=5+5^3+5^5+\cdots+5^{101}\)

=>\(25F=5^3+5^5+5^7+\cdots+5^{103}\)

=>\(25F-F=5^3+5^5+5^7+\cdots+5^{103}-5-5^3-5^5-\cdots-5^{101}\)

=>\(24F=5^{103}-5\)

=>\(F=\frac{5^{103}-5}{24}\)

Ta có: \(5^{103}+1>5^{103}-5\)

=>\(\frac{5^{103}+1}{24}>\frac{5^{103}-5}{24}\)

=>E>F