a) \(\left(5x+3^4\right).6^8=6^9.3^4\)

\(=>6x+3^4=3^4.6^9:6^8\)

\(=>6x+3^4=3^4.6\)

\(=>6x=6.3^4-3^4\)

\(=>6x=0\)

\(=>x=0:6\)

\(=>x=0\)

a/(5x + 3⁴).6⁸=6⁹.3⁴

   (5x + 3⁴) = 6⁹:6⁸.3⁴

   5x + 81 = 6.81

   5x = 6.81 - 81

   5x = 486 - 81

   5x = 425

   x = 425:5

   x = 85

b/92 - 2x = 2.4²- 3.4 + 120:15
   92 - 2x = 2.16 - 12 + 8

   92 - 2x = 32 - 12 + 8

   92 - 2x = 28

   2x = 92 - 28

   2x = 64

   x = 64:2

   x = 32

c/5³.(3x + 2) : 13 = 10³: (13⁵:13⁴)

  125.(3x + 2) : 13 = 1000:13 

  125.(3x+2) = 1000:13.13

  125.(3x+2) = 1000

   3x + 2 = 1000:125

   3x + 2 = 8

   3x = 8 - 2

   3x = 6

   x = 6:3

   x = 2

Bạn nhớ tick cho mình nhé!

a/(5x + 34).68=69.34

   (5x + 34) = 69:68.34

   5x + 81 = 6.81

   5x = 6.81 - 81

   5x = 486 - 81

   5x = 405

   x = 405:5

   x = 81

b/92 - 2x = 2.42- 3.4 + 120:15
   92 - 2x = 2.16 - 12 + 8

   92 - 2x = 32 - 12 + 8

   92 - 2x = 28

   2x = 92 - 28

   2x = 64

   x = 64:2

   x = 32

c/53.(3x + 2) : 13 = 103: (135:134)

  125.(3x + 2) : 13 = 1000:13 

  125.(3x+2) = 1000:13.13

  125.(3x+2) = 1000

   3x + 2 = 1000:125

   3x + 2 = 8

   3x = 8 - 2

   3x = 6

   x = 6:3

   x = 2

Cái này mới đúng

Bạn nhớ tick cho mình nhé!

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}\)

\(A=\frac{2019}{2020}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(2B=1-\frac{1}{2019}\)

\(2B=\frac{2018}{2019}\)

\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)

\(C=3^0+3^1+3^2+...+3^{100}\)

\(3C=3^1+3^2+3^3+...+3^{101}\)

\(3C-C=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(1+3^1+3^2+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)

a) ( x - 4 ) . ( x - 3 ) = 0

Thỏa mãn điều kiện x - 4 hoặc x - 3 bằng 0 .

\(\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}\)

b) ( x - 4 ) . ( y - 3 ) = 0

Thỏa mãn điều kiện x - 4 hoặc y - 3 bằng 0 .

\(\Rightarrow\hept{\begin{cases}x=4,x=3\\y=4,y=3\end{cases}}\)

c) x - 24 : 12 = 15

x - 2 = 15

x = 13

d) ( x - 12 ) : 13 = 5

x - 12 = 65

x = 77

e) 1 + 3 + 5 + ..... + x = 36

..... + x = 27

..... + x = 20

x = 11

f) 2 + 4 + .... + 2x = 110

2 + 4 + ....x = 55

Thỏa mãn điều kiện \(x\in N\left|2x+0\right|x⋮2\)

\(x\in\varnothing\)

`@`Thay `x=2` vào `A` có:

 `A=3^2-9.2=9-18=-9`

`@` Thay `x=1/3` vào `A` có:

`A=(1/3)^2-9. 1/3=1/9-3=-26/9`

Khi x=2 thì \(A=3\cdot2^2-9\cdot2=12-18=-6\)

Khi x=1/3 thì \(A=3\cdot\dfrac{1}{9}-9\cdot\dfrac{1}{3}=\dfrac{1}{3}-3=-\dfrac{8}{3}\)

a).\(541+\left(218-x\right)=735\)

                        \(218-x=735-541\)

                       \(218-x=194\)

                                      \(x=218-194\)

                                       \(x=24\)

b).\(5\left(x+35\right)=515\)

         \(x+35=515:5\)

          \(x+35=103\)

         \(x=103-35\)

          \(x=68\)

c).\(96-3\left(x+1\right)=42\)

                 \(3\left(x+1\right)=96-42\)

                   \(3\left(x+1\right)=54\)

                      \(\left(x+1\right)=54:3\)

                      \(\left(x+1\right)=18\)

                        \(x=18-1\)   

                        \(x=17\)

d)\(12x-33=3^2\cdot3^3\)

   \(12x-33=3^5\)

   \(12x-33=243\)

                \(12x=243+33\)

                \(12x=276\)

                     \(x=276:12\)

                     \(x=23\)

\(Nhớ\)\(tk\)\(mình\)\(nha\)\(!\)

A=1+3+3^2+3^3+3^4+...+3^100

3A=3+3^2+3^3+3^4+...+3^101

3A-A=(3+3^2+3^3+3^4+...+3^101)-(1+3+3^2+3^3+3^4+...+3^100)

2A=3^101-1

A=(3^101-1):2

phần b làm tương tự phần a nhưng mà là nhân cả biểu thức B với 4 nhé

a)   \(x^3=216\)

=>  \(x^3=6^3\)

=>  \(x=6\)

ta có :