a) 7.x - x = 5²¹ : 5¹⁹ + 3.2² - 7⁰

6x = 25 + 12 - 1

6x = 36

x = 6

b) 7x - 2x = 6¹⁷ : 6¹⁵ + 44 : 11

5x = 36 + 4

5x = 40

x = 8

c) 5x + x = 39 - 3¹¹ : 3⁹

6x = 39 - 9

6x = 30

x = 5

d) [(6x - 39) : 7]. 4 = 12

(6x - 39) : 7 = 12 : 4

(6x - 39) : 7 = 3

6x - 39 = 3 . 7

6x - 39 = 21

6x = 21 + 39

6x = 60

x = 10

x = 

\(32^{15}=\left(2^5\right)^{15}=2^{5.15}=2^{75}\)

\(4^{39}=\left(2^2\right)^{39}=2^{2.39}=2^{78}\)

Do \(2^{78}>2^{75}\)

\(\Rightarrow4^{39}>32^{15}\)

\(\Rightarrow1+4+4^2+...+4^{39}>32^{15}\)

\(\Rightarrow3\left(1+4+4^2+...+4^{39}\right)>32^{15}\)

Vậy \(A>B\)

mọi ng giúp mik với

a) \(2^{4x}-3^{2x}=144-164:41\)

\(\Leftrightarrow16^x-9^x=144-4\)

\(\Leftrightarrow\) ......

bjan tự tính tiếp đc òi                 

a, Bội (6) = {0; 6}

b, Số đối của: -4 = 4 ; 0 = 0

c, \(3^2+10:2=9+10:2=9+5=14\)

Câu 2:

\(\left(15-\left[3^{20}:3^{19}+2022^0\right]\right):11=\left(15-\left[3^{20-19}+1\right]\right):11=\left(15-\left[3^1+1\right]\right):11\)

\(=\left(15-4\right):11=11:11=1\)

Câu 3:

\(2x-7=39\)

\(2x=39+7\)

\(2x=46\)

\(x=46:2\)

\(x=23\)

=0 bạn nha

a,(\(6x-5x^2-15+2x^3:\left(2x-5\right)\)

\(\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)\)

a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

Bài 1: 

a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)

\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

b) Ta có: \(\left(2x-3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)

\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Bài 2: 

a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)

b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)

c) \(3+3^2+3^3+...+3^{2007}\)

\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2005}\right)⋮13\)

a) \(\left(x-1\right)^2=49\)

\(\Rightarrow\left(x-1\right)^2=7^2=\left(-7\right)^2\)

\(\Rightarrow x-1=7\)                        hoặc                           \(x-1=-7\)

\(x=7+1=8\)                                                                \(x=-7+1=-6\)

Vậy x = 8 hoặc x = - 6

b) \(3\cdot\left(13-x\right)^2=27\)

\(\left(13-x\right)^2=27\div3=9\)

\(\Rightarrow\left(13-x\right)^2=3^2=\left(-3\right)^2\)

\(\Rightarrow13-x=3\)                                          hoặc                                    \(13-x=-3\)

\(x=13-3=10\)                                                                                            \(x=13+3=16\)

Vậy x = 10 hoặc x = 16

c) \(164-\left(15-x\right)^3=100\)

\(\left(15-x\right)^3=164-100=64\)

\(\Rightarrow\left(15-x\right)^3=4^3\)

\(\Rightarrow15-x=4\)

\(x=15-4=11\)

Vậy x = 11

d) \(\left(x+3\right)^3-15=210\)

\(\left(x+3\right)^3=210+15=225\)

\(\Rightarrow\left(x+3\right)^3=...\)

Tương tự mũ lẻ cậu nhé

e) \(x^2\div4=16\)

\(x^2=16\cdot4=64\)

\(\Rightarrow x^2=8^2=\left(-8\right)^2\)
Vậy x = 8 hoặc x = - 8

a/\(\left(x-1\right)^2\)=49

\(\left(x-1\right)^2\)=\(7^2\)

=>x-1=7

x=7+1

x=8

b/3.\(\left(13-x\right)^2\)=27

\(\left(13-x\right)^2\)=27:3

\(\left(13-x\right)^2\)=9

\(\left(13-x\right)^2\)=\(3^2\)

=>13-x=3

x=13-3

x=10

c/164-\(\left(15-x\right)^3\)=100

\(\left(15-x\right)^3\)=164-100

\(\left(15-x\right)^3\)=64

\(\left(15-x\right)^3\)=\(4^3\)

=>15-x=4

x=15-4

x=11

d/\(\left(x+3\right)^3\)-15=210

\(\left(x+3\right)^3\)=210+15

\(\left(x+3\right)^3\)=225

sai đề bài câu d hay sao ý bạn ạ 

chỉ có \(\left(x+3\right)^2\)thì mới tính được

e/\(x^2\):4=16

\(x^2\)=16.4

\(x^2\)=64

\(x^2\)=\(8^2\)

=>x=8

 Bài 1:

1; 5\(x\) + \(x\) = 39 - 3¹¹ : 3⁹

    \(x\).(5 + 1) = 39 - 3²

    \(x.6\)        = 39 - 9

   \(x.6\)         = 30

   \(x\)           = 30 : 6

   \(x\)          = 5

Vậy \(x\) = 5

2; 5\(x\) + \(x\) = 150 : 2 + 3

    \(x\).(5 + 1) = 75 + 3

    \(x.6\)          = 78

    \(x\)             = 78 : 6

     \(x\)            = 13

Vậy \(x=13\)