S=3⁰+3²+3⁴+3⁶+...+3²⁰⁰

6S=3²+3⁴+3⁶+...+3²⁰²

6S-S=(3²+3⁴+3⁶+...+3²⁰²)-(1+3²+3⁴+...+3²⁰⁰)

5S=1+(3²-3²)+(3⁴-3⁴)+...+(3²⁰⁰-3²⁰⁰)+3²⁰²

S=(3²⁰⁰+1):5\(\frac{ }{ }\)

a)x+3/6=7/2

x       =7/2-3/6

x     =3

b)2,8+(3x-5/6)=2/2/5

  14/5+(3x-5/6)=12/5

          (3x-5/6)=12/5-14/5

         3x+5/6=-2/5

        3x         =-2/5-5/6

       3x          =-37/30

          x         =-37/36:3

         x          =-37/90

CẢM ƠN PHAN THAO

Bài 1:

6) 3x + 2³ = 17 + 3²

3x + 8 = 17 + 9

3x + 8 = 26

3x = 26 - 8

3x = 18

x = 18 : 3

x = 6

Vậy x = 6

Bài 2:

3) 145 - (125 + x) = 12

125 + x = 145 - 12

125 + x = 133

x = 133 - 125

x = 8

Vậy x = 8

6) 3³ - (x - 5) = 2²

27 - (x - 5) = 4

x - 5 = 27 - 4

x - 5 = 23

x = 23 + 5

x = 28

Vậy x = 28

9) (x + 7) - 15⁰ = 202 - 19

(x + 7) - 1 = 189

x + 7 = 189 + 1

x + 7 = 190

x = 190 - 7

x - 183

Vậy x = 183

a) 2 . x - 138 = 2³ . 3²

2 . x - 138 = 8 . 9

2 . x - 138 = 72

2 . x = 72 + 138

2 . x = 210

     x = 210 : 2

     x = 105

b) 231 - ( x - 6 ) = 1339 : 13

231 - ( x - 6 ) = 103

x - 6 = 231 - 103

x - 6 = 128

     x = 128 + 6

     x = 134

a, 2x-138=32.9

2x-138=288

2x=288+138

2x=426

x=213

b, 231-(x-6)=103

x-6=231-103

x-6=128

x=134

Bạn chỉ mình cách viết hỗn số đi rồi mình sẽ chỉ bài cho

\(2\frac{5}{6}.x-\frac{3}{2}=3\frac{5}{6}\)

\(\Rightarrow\)\(2\frac{5}{6}.x=3\frac{5}{6}+\frac{3}{2}\)

\(\Rightarrow\)\(2\frac{5}{6}.x=\frac{16}{3}\)

\(\Rightarrow\)\(x=\frac{16}{3}:2\frac{5}{6}\)

\(\Rightarrow\)\(x=\frac{32}{17}\)

a) 25 : x = x

25 = x^2

x^2 = ( +-5 )^2

b) 3358 : 23 = 2x - 6

146 = 2x - 6

2x = 152

x = 76

c) ( 2x + 1 )^3 = 27 = 3^3

=> 2x + 1 = 3

=> 2x = 2

=> x = 1

d) ( x - 2 )^3 = ( x - 2 )^2

( x - 2 )^2 . ( x - 2 ) - ( x -2 )^2 = 0

( x - 2 )^2 . [ ( x - 2 ) - 1 ] = 0

+) x - 2 = 0

=> x = 2

+) x - 2 - 1 = 0

x - 3 = 0

x = 3

\(25\div x=x\Rightarrow x.x=25\Rightarrow x^2=25\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

\(3358\div23=2x-6\)

\(\Rightarrow2x-6=146\)

\(\Rightarrow2x=152\)

\(\Rightarrow x=\frac{152}{2}=76\)

\(\left(2x+1\right)^3=27\)

Mà \(3^3=27\)

Nên \(\left(2x+1\right)^3=3^3\)

\(\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)

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

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

\(\Rightarrow\left(x-2\right)^2.\left(x-2-1\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

Vậy......................

76