\(A=3+3^2+3^3+3^4+.....+3^{25}+3^{26}\)

\(\Rightarrow A+1=1+3+3^2+3^3+3^4+....+3^{25}+3^{26}\)

\(\Rightarrow A+1=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+.....+\left(3^{24}+3^{25}+3^{26}\right)\)

\(A+1=40+3^3\left(1+3+3^2\right)+....+3^{24}\left(1+3+3^2\right)\)

\(A+1=40+3^3\cdot40+....+3^{24}\cdot40\)

\(A+1=40\left(1+3^3+...+3^{24}\right)\)

\(\Rightarrow\left(A+1\right)⋮40\)

\(\Rightarrow A:40\)dư 39

dư 3 mới đúng bạn ơi

A= 3+3^2+3^3+...+3^25 (25 SH)

  =3^2+(3+3^3)+(3^4+3^6)+...+(3^23+3^25)

  =9+40.1+3^3.(3+3^3)+...+3^22.(3+3^3)

  =9+40.1+3^3.40+...+3^22.40

  =9+40.(1+3^3+...+3^22) chia 40 dư 9

(1+3^3+....+3^22 thuộc N ; 40 chia hết cho 40 nhưng 9 không chia hết cho 40)

Vậy Achia 40 dư 9

nhớ k đúng nhe!

128 - 3.95 - 2\(x\) = 107

128 -  285 - 2\(x\)   =107

-157 - 2\(x\)           = 107

          2\(x\)          = -107 - 157 

          2\(x\)          = -264

            \(x\)          = -264 : 2

            \(x\)          = -132

b, (3\(x\) - 25) - (\(x\) - 9) = 2 - \(x\) 

     3\(x\) - 25  - \(x\) + 9  = 2  - \(x\)

    3\(x\) - \(x\) + \(x\) = 2 + 25 - 9

     3\(x\)           = 18

        \(x\)          = 18 : 3

        \(x\)        = 6

1) \(\left|2x+5\right|\ge21\Rightarrow2x+5\ge21\)hoặc \(2x+5

2b) Áp dụng bất đẳng thức giá trị tuyệt đối: |a| + |b|  \(\ge\) |a + b|. Dấu "=" xảy ra khi tích a.b \(\ge\) 0 

Ta có: B = |2x - 1| + |3 - 2x| + 5  \(\ge\) |2x - 1+3 - 2x| + 5  = |2| + 5 = 7

=> Min B = 7 khi

(2x - 1)( 3 - 2x) \(\ge\) 0 => (2x - 1)(2x - 3) \(\le\) 0 

Mà 2x - 1 > 2x - 3 nên 2x - 1 \(\ge\) 0 và 2x - 3 \(\le\)  0 

=> x \(\ge\) 1/2 và x  \(\le\) 3/2

a) 2/3 + 1/3(2x-1) =-5

            1/3(2x-1) = -5 - 2/3

            1/3(2x-1) = -17/3

                 2x-1  = -17/3 : 1/3 

                 2x-1  = -17

                 2x     = -17 + 1 

                 2x     = -16 

                  x      = -16 : 2 

                  x      = -8  

b) 25% .x + 2x = 4 

   x .( 25%+2 ) = 4

   x . 9/4          = 4 

   x                  = 4 : 9/4 

   x                  = 16/9

a) \(\left(2x+3\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+3\right)^2=\left(\frac{3}{5}\right)^2\) hoặc \(\left(2x+3\right)^2=\left(-\frac{3}{5}\right)^2\)

\(\Rightarrow2x+3=\frac{3}{5}\) hoặc \(2x+3=-\frac{3}{5}\)

\(\Rightarrow2x=-\frac{12}{5}\) hoặc \(2x=-\frac{18}{5}\)

\(\Rightarrow x=-\frac{6}{5}\) hoặc \(x=-\frac{9}{5}\)

Vậy.......

b) \(\left(3x-1\right)^3=\frac{1}{27}\)

\(\left(3x-1\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Rightarrow3x-1=\frac{1}{3}\)

\(\Rightarrow3x=\frac{4}{3}\)

\(\Rightarrow x=\frac{4}{9}\)

Vậy.......

a) (2x+3)²=\(\frac{9}{25}\)

2x+3=\(\sqrt{\frac{9}{25}}\)=\(\frac{3}{5}\)

2x=\(\frac{-12}{5}\)

x=\(\frac{-6}{5}\)

b) (3x-1)³=\(\frac{-1}{27}\)

3x-1=\(\frac{-1}{3}\)

3x=\(\frac{2}{3}\)

x=\(\frac{2}{9}\)

a) ( 2x + 3 )² = \(\frac{9}{25}\)

\(\Rightarrow2x+3=\frac{3}{5}\)hoặc \(-\frac{3}{5}\)

\(\Rightarrow2x=-\frac{12}{5}\)hoặc \(-\frac{18}{5}\)

\(\Rightarrow x=-\frac{6}{5}\)hoặc\(-\frac{9}{5}\)

b) ( 3x - 1)³ = \(-\frac{1}{27}\)

\(\Rightarrow3x-1=-\frac{1}{3}\)

\(\Rightarrow3x=\frac{2}{3}\)

\(\Rightarrow x=\frac{2}{9}\)

 A, 2 . ( x - 3 ) + 1 = -25 

2.(x-3)=-25-1

2.(x-3)=-26

x-3    =-26:2

x-3    =-13

x=-13+3

x=-10

 B , | 3 . x + 1 | + 24 = 13

|3.x+1|=13-24

|3.x+1|=-11

TH1:

3.x+1=-11

3.x    =-11-1

3.x   =-12

x      =-12:3

x      =-4

TH2:

3.x+1=-(-11)

3.x+1=11

3.x    =11-1

3x=10

x=10/3

vậy x=-4 ;10/3

 a), 2 . ( x - 3 ) + 1 = -25 

2.(x-3)=-25-1

2.(x-3)=-26

x-3    =-26:2

x-3    =-13

x=-13+3

x=-10

  b)| 3 . x + 1 | + 24 = 13

|3.x+1|=13-24

|3.x+1|=-11

TH1:

3.x+1=-11

3.x    =-11-1

3.x   =-12

x      =-12:3

x      =-4

TH2:

3.x+1=-(-11)

3.x+1=11

3.x    =11-1

3x=10

x=10/3

vậy x=-4 ;10/3