\(78+23\cdot81-69=78+1863-69\)

                                   \(=1941-69\)

                                   \(=1872\)

\(78+23\cdot\left(81-69\right)=78+23\cdot12\)

                                        \(=78+276\)

                                        \(=354\)

\(\left(78+23\right)\cdot81-69=101\cdot81-69\)

                                        \(=8181-69\)

                                        \(=8112\)

\(\left(78+23\right)\cdot\left(81-69\right)=101\cdot12\)

                                              \(=1212\)

1872

354

8112

1212

a) 2³.(20 - 2x) = 4.2⁵

8.(20 - 2x) = 4.32

8.(20 - 2x) = 128

20 - 2x = 128 : 8

20 - 2x = 16

2x = 20 - 16

2x = 4

x = 4 : 2

x = 2

b) 3²⁺ˣ = 81

3²⁺ˣ = 3⁴

2 + x = 4

x = 4 - 2

x = 2

a)

\(\left[\left(x-81\right)^3:5^3\right]-2^3=0\\ =>\left(x-81\right)^3:5^3=2^3\\ =>\left(x-81\right)^3=2^3.5^3=10^3\\ =>x-81=10\\ =>x=91\)

b) 

\(3^{n+1}.3^{n+3}=18^{10}:6^{10}\\ =>3^{n+1+n+3}=3^{10}\\ =>2n+4=10\\ =>2n=6=>n=3\)

a) \(\left[\left(x+81\right)^3:5^3\right]-2^3=0\)

\(\Rightarrow\left(\dfrac{x+81}{5}\right)^3=2^3\)

\(\Rightarrow\dfrac{x-81}{5}=2\)

\(\Rightarrow x-81=10\)

\(\Rightarrow x=91\)

b) \(3^{n+1}\cdot3^{n+3}=18^{10}:6^{10}\)

\(\Rightarrow3^{2n+4}=3^{10}\)

\(\Rightarrow2n+4=10\)

\(\Rightarrow2n=6\)

\(\Rightarrow n=3\)

a 54

b 41

k cho minh nhe ket ban khong

\(a,123+x=23\times3\)

    \(123+x=69\)

                 \(x=69-123\)

                 \(x=-54\)

a,123+x=69                                 

x=69-123

x=-54

B,\(2x^3=2^3\)

Suy ra:

x=2:2

x=1

=2/7x7/2x23/81x81/23

=1x1=1

2 / 7 X 23 / 81 X 7/2 X 81/23

= ( 2/7 X 7/2 ) X ( 23 / 81 X 81 / 23 )

= 1 X 1

= 1 

k mk vs nhé <3

\(\frac{2}{7}\times\frac{23}{81}\times\frac{7}{2}\times\frac{81}{23}=1\)

Các tử số và mẫu số giống nhau đều bị gạch bỏ.

Bài 2:

a: \(17-x=3\)

=>\(x=17-3\)

=>x=14(nhận)

b: \(2\cdot\left(x-1\right):3=6\)

=>\(2\left(x-1\right)=6\cdot3=18\)

=>x-1=18/2=9

=>x=9+1=10(nhận)

c: \(x+\left(-2\right)=\left(-11\right)+7\)

=>\(x-2=-4\)

=>\(x=-4+2=-2\left(loại\right)\)

d: \(\left(x-1\right)^2-5=20\)

=>\(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Câu 3:

a: Đặt *=a

\(\overline{57a3}⋮9\)

=>\(5+7+a+3⋮9\)

=>\(a+15⋮3\)

mà 0<=a<=9

nên a=3

=>*=a

b: \(A=123\cdot7+8+9\)

123*7 là số lẻ

9 là số lẻ

=>123*7+9 chia hết cho 2

mà 8 chia hết cho 2

nên \(A=123\cdot7+9+8⋮2\)

\(123\cdot7⋮3;9⋮3;8⋮̸3\)

=>\(A=123\cdot7+9+8⋮̸3\)

c: \(B=3\cdot5\cdot7+10^{50}\)

\(=5\cdot3\cdot7+5\cdot5^{49}\cdot2^{49}\)

\(=5\left(3\cdot7+5^{49}\cdot2^{49}\right)⋮5\)

=>B là hợp số

a,  3 x + 1 : 3 4 = 81

3 x - 3 = 3 4

x – 3 = 4

x = 7

Vậy x = 7

b,  3 x + 3 . 3 x + 1 = 729

3 2 x + 4 = 3 6

2x + 4 = 6

x = 1

Vậy x = 1

c,  2 x + 3 . 2 x = 128

2 2 x + 3 = 2 7

2x + 3 = 7

x = 2

Vậy x = 2

d,  23 + 3 x = 5 6 : 5 3

23 + 3 x = 5 3

23 + 3x = 125

3x = 102

x = 34

Vậy x = 34

e,  2 x + 2 x + 4 = 272

2 x + 2 x . 2 4 = 272

2 x ( 1 + 2 4 ) = 272

2 x . 17 = 272

2 x = 16

2 x = 2 4

x = 4

Vậy x = 4

\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)

\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)

\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)

\(\Leftrightarrow X=\frac{109}{6075}\)

Vậy X=109/6075

Chắc Sai kết quả chứ công thức đúng nha!!!...

Fighting!!!...

Đặt: 

 \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)

=> \(A=\frac{12}{25}\)

Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

   \(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)

=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)

Giải phương trình:

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)

                        \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)

                                                                            \(12x+\frac{12}{25}=11x+\frac{121}{243}\)

                                                                             \(12x-11x=\frac{121}{243}-\frac{12}{25}\)

                                                                                                  \(x=\frac{109}{6075}\)