\(A=3+3^2+3^3+...+3^{2023}\)

\(\Rightarrow A+1=1+3+3^2+3^3+...+3^{2023}\)

\(\Rightarrow A+1=\dfrac{3^{2023+1}-1}{3-1}=\dfrac{3^{2024}-1}{2}\)

\(\Rightarrow A=\dfrac{3^{2024}-1}{2}-1=\dfrac{3^{2024}-3}{2}=\dfrac{3\left(3^{2023}-1\right)}{2}\)

\(x\) - 5 = \(x\) + 5

\(x\) - \(x\) = 5 + 5

 0 = 10 (vô lý)

vậy \(x\in\) \(\varnothing\)

10=2x5

12=2²x3

20=2²x5

25=5²

42=2x3x7

60=2²x3x5

100=2²x5²

124=2²x31

240=2⁴x3x5.

10= 2 x 5

12= 2² x 3

20 = 2² x 5

25=5²

42= 2 x 3 x 7

60 = 2² x 3 x 5

100= 2² x 5²

124 = 2² x 31

240 = 2⁴ x 15 

\(A=\dfrac{2x+5}{x+1}\left(x\ne-1\right)\)

Để A nguyên khi và chỉ khi :

\(2x+5⋮x+1\)

\(\Rightarrow2x+5-2\left(x+1\right)⋮x+1\)

\(\Rightarrow2x+5-2x-2⋮x+1\)

\(\Rightarrow3⋮x+1\)

\(\Rightarrow x+1\in\left\{-1;1;-3;3\right\}\)

\(\Rightarrow x\in\left\{-2;0;-4;2\right\}\left(x\in Z;x\ne-1\right)\)

uuuu

(x+1)+(x+3)+...+(x+99)=1000

=>50x+1+3+...+99=1000

=>50x+\(\dfrac{\left(99-1\right):2+1\cdot\left(1+99\right)}{2}\)=1000

=>50x+2500=1000

=>50x=1000-2500=-1500

=>x=-1500:50=-30.

\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=1000\)

\(\left(x+x+x+...+x\right)+\left(1+3+5+7+9+...+99\right)\)

\(\left(x.50\right)+2500\)=1000

\(x.50=1000-2500\)

\(x.50=-1500\)

\(x=-1500:50=-30\)

674355

3+6+9+...+2010=\(\dfrac{\left(2010-3\right):3+1\cdot\left(3+2010\right)}{2}\)=674355.

1+2+3+...+X=500500

=>\(\dfrac{\left(x-1\right)+1\cdot\left(1+x\right)}{2}\)=500500

=>x\(\cdot\)(x+1)=500500x2=1001000=1000x1001

=>x=1000.

\(\text{∘ Ans}\)

\(\downarrow\)

`(x - 2).(x - 3) = 0`

`\Rightarrow`

`\text {TH1: } x - 2 = 0`

`\Rightarrow x = 0 + 2`

`\Rightarrow x = 2`

`\text {TH2: } x - 3 = 0`

`\Rightarrow x = 0 + 3`

`\Rightarrow x = 3`

Vậy, `x \in`\(\left\{2;3\right\}\)

(x-2)(x-3)=0

=>x-2=0 hoặc x-3=0

=>x=2 hoặc x=3.

64x25+23x25+25=25(64+23+1)=25x88=2200

b,58x42+32x8+5x16=58x42+16x16+5x16

=58x42+16(16+5)=58x42+16x21=58x42+8x42=42(58+8)=42x56=2352.

1673

2604

Gọi mẫu số là \(x\);  \(x\) \(\ne\) 0; \(x\in\) Z thì tử số là \(x+6\)

Theo bài ra ta có: \(\dfrac{x+6}{x}\)   = \(\dfrac{11}{3}\)

                             3\(x\) + 18 = 11\(x\) 

                              11\(x\) - 3\(x\) = 18

                               8\(x\)         = 18

                                  \(x\)          = \(\dfrac{18}{8}\) 

                                     \(x\) = \(\dfrac{9}{4}\) (loại)

 Không có phân số nào thỏa mãn đề bài