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 

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

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:

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

\(\dfrac{15}{34}+\dfrac{1}{3}+\dfrac{19}{34}-\dfrac{4}{3}+\dfrac{3}{7}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{3}{7}=1-1+\dfrac{3}{7}=\dfrac{3}{7}\)

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

a: \(640=2^7\cdot5;1440=2^5\cdot3^2\cdot5\)

=>ƯCLN(640;1440)\(=2^5\cdot5=32\cdot5=160\)

640⋮a; 1440⋮a

=>a∈ ƯC(640;1440)

mà a lớn nhất

nên a=ƯCLN(640;1440)=160

b: \(450=2\cdot3^2\cdot5^2;210=2\cdot3\cdot5\cdot7\)

=>ƯCLN(450;210)\(=2\cdot3\cdot5=30\)

450⋮a; 210⋮a

=>a∈ ƯC(450;210)

mà a lớn nhất

nên a=ƯCLN(450;210)=30

c: \(128=2^7;210=2\cdot3\cdot5\cdot7\)

=>ƯCLN(128;210)=2

128⋮a; 210⋮a

=>a∈ ƯC(128;210)

=>a∈ Ư(2)

mà 6<a<15

nên a∈∅

d: \(2350=2\cdot5^2\cdot47;1260=2^2\cdot3^2\cdot5\cdot7\)

=>ƯCLN(2350;1260)=\(2\cdot5=10\)

2350⋮a; 1260⋮a

=>a ∈ƯC(2350;1260)

=>a∈ Ư(10)

mà 80<a<140

nên a∈∅

e: \(112=2^4\cdot7;140=2^2\cdot5\cdot7\)

=>ƯCLN(112;140)\(=2^2\cdot7=28\)

112⋮a; 140⋮a

=>a ∈ƯC(112;140)

=>a∈ Ư(28)

mà 10<a<20

nên a=14

f: \(144=2^4\cdot3^2;192=2^6\cdot3\)

=>ƯCLN(144;192)\(=2^4\cdot3=16\cdot3=48\)

144⋮a; 192⋮a

=>a∈ ƯC(144;192)

=>a∈ Ư(48)

mà a<20

nên a∈{1;2;3;4;6;8;12;16}

g: \(420=2^2\cdot3\cdot5\cdot7;700=2^2\cdot5^2\cdot7\)

=>ƯCLN(420;700)\(=2^2\cdot5\cdot7=4\cdot5\cdot7=20\cdot7=140\)

420⋮a; 700⋮a

=>a∈ ƯC(420;700)

=>a∈ Ư(140)

mà a lớn nhất

nên a=140

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

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

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.

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

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