\(a,70-5\times\left(x-3\right)=45\)

\(5\times\left(x-3\right)=70-45\)

\(5\times\left(x-3\right)=25\)

\(x-3=25:5\)

x - 3 = 5

x = 5 + 3 = 8

vậy x = 8

\(b,12+\left(5+x\right)=20\)

12 + 5 + x = 20

17 + x = 20

x = 20 - 17 = 3

vậy x = 3

c, 130 - (100 + x) = 25

130 - 100 - x = 25

30 - x = 25

x = 30 - 25 = 5

vậy x = 5

\(g,10+2\times x=4^5:4^3\)

\(10+2\times x=4^2\)

\(10+2\times x=16\)

\(2\times x=16-10\)

\(2\times x=6\)

x = 6 : 2 = 3

vậy x = 3

h, \(14\times x+54=82\)

\(14\times x=82-54\)

\(14\times x=28\)

\(x=28:14=2\)

vậy x = 2

\(k,15\times x-133=17\)

\(15\times x=17+133\)

\(15\times x=150\)

\(x=150:15=10\)

vậy x = 10

\(-28-7.|-3\chi+15|=-70\)

\(\Rightarrow7.|-3\chi+15|=42\)

\(\Rightarrow|-3\chi+15|=6\)

\(\Rightarrow\orbr{\begin{cases}-3\chi+15=6\\-3\chi+15=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-3\chi=-9\\-3\chi=-21\end{cases}}\Rightarrow\orbr{\begin{cases}\chi=3\\\chi=7\end{cases}}\)

HTDT

\(c,12-2\left(-\chi+3\right)^2=-38\)

\(\Rightarrow2\left(-\chi+3\right)^2=50\)

\(\Rightarrow\left(-\chi-3\right)^2=25\)

\(\Rightarrow\left(-\chi-3\right)^2=\left(\pm5\right)^2\)

\(\Rightarrow\orbr{\begin{cases}-\chi-3=5\\-\chi-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}\chi=-8\\\chi=-2\end{cases}}\)

HTDT

yêu cầu của đề bài là gì vậy bạn

A = \(\left(\frac{1}{11}+\frac{1}{12}+.........+\frac{1}{20}\right)\)  +  \(\left(\frac{1}{21}+\frac{1}{22}+..........+\frac{1}{30}\right)\)+ \(\left(\frac{1}{31}+.....+\frac{1}{60}\right)\)+ ... + \(\frac{1}{70}\)

Nhận xét: 

\(\frac{1}{11}\)+ \(\frac{1}{12}\)+ ........  +  \(\frac{1}{20}\)> \(\frac{1}{20}\)+\(\frac{1}{20}\)+........+\(\frac{1}{20}\)> \(\frac{10}{20}\)>\(\frac{1}{2}\)

\(\frac{1}{21}+\frac{1}{22}+.......+\frac{1}{30}>\frac{30}{60}>\frac{1}{2}\)

\(\frac{1}{31}+......+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+.......+\frac{1}{60}>\frac{30}{60}>\frac{1}{2}\)

A > \(\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+......+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}>\frac{4}{3}\)

Dạng 3 :

a) 3x - 10 = 2x + 13

=> 3x - 2x = 13 - 10

=> x = 3

b) x + 12 = -5 - x

=> x + x = -5 - 12

=> 2x = -17

=> x = -8,5

c) x + 5 = 10 - x 

=> x + x = 10 - 5

=> 2x = 5

=> x = 2,5

d) 6x + 23 = 2x - 12

=> 2x - 6x = 23 + 12

=> -4x = 35

=> x = -8,75

e) 12 - x = x + 1

=> x + x = 12 - 1

=> 2x = 11

=> x = 5,5

f) 14 + 4x = 3x + 20

=> 4x - 3x = 20 - 14

=> x = 6

***********************************************************

a) Ta có: \(\dfrac{x+12}{10-x}=-\dfrac{x-10+22}{x-10}=-1+\dfrac{22}{x-10}\)

Vì \(\left(x+12\right)⋮\left(10-x\right)\) nên \(22⋮\left(x-10\right)\)

Do đó ta có bảng:

Vậy \(x\in\left\{-12;-1;8;9;11;12;32\right\}\)

c) \(\left(x-3\right)⋮\left(x+1\right)\)

=> \(\left(x-3\right)-\left(x+1\right)⋮\left(x+1\right)\)

=> \(\left(x-3-x-1\right)⋮\left(x+1\right)\)

=>\(-4⋮\left(x+1\right)\)

=> x+1\(\in\) ư(-4)= \(\left\{\pm1,\pm2,\pm4\right\}\)

ta có bảng sau

vậy x\(\in\left\{-5,-3;-2;0;1;3\right\}\)

Bài 1: 

a) \(33^{2x}:11^{2x}=81\)\(\Leftrightarrow\left(33:11\right)^{2x}=81\)

\(\Leftrightarrow3^{2x}=3^4\)\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

b) \(\frac{x}{-5}=\frac{4}{21}\)\(\Leftrightarrow21x=-20\)\(\Leftrightarrow x=\frac{-20}{21}\)

Vậy \(x=\frac{-20}{21}\)

Bài 2:

\(A=\frac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)

\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+\left(3^2+3^6+3^{10}+3^{14}\right)}\)

\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+3^2.\left(1+3^4+3^8+3^{12}\right)}\)

\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right).\left(1+3^2\right)}=\frac{1}{1+3^2}=\frac{1}{1+9}=\frac{1}{10}\)

\(33^{2x}:11^{2x}=81\)!

\(\left(33:11\right)^{2x}=81\)

\(3^{2x}=81\)

\(3^{2x}=3^4\)

\(2x=4\)

\(x=4:2\)

\(x=2\)

vậy \(x=2\)

\(\frac{x}{-5}=\frac{4}{21}\)

x.21=-5.4

x.21=-20

x=-20:21

\(x=-\frac{20}{21}\)

vậy \(x=-\frac{20}{21}\)