a: \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)

=>\(9x^2+12x+4-\left(9x^2-12x+4\right)-5x-38=0\)

=>\(9x^2+7x-34-9x^2+12x-4=0\)

=>19x-38=0

=>19x=38

=>x=38/19=2

b: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

=>\(x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)

=>\(x^3+3x^2+12x-9=x^3+3x^2+3x+1\)

=>12x-9=3x+1

=>12x-3x=1+9

=>9x=10

=>x=10/9

a) x.(x+1) = (2+2500)[(2500-2):2+1]:2 =2502.625=1563750

  x(x+1) = 1250(1250+1) 

=> x = 1250

b) 1+2+3+4+.......+x = 820

 (1+x)x:2 =820

x(x+1) = 1640 =40(40+1)

=> x = 40

`123-x=38`

`=> x= 123-38`

`=>x=85`

Vậy `x=85`

__

`(x+15) -7=33`

`=>x+15=33+7`

`=>x+16= 40`

`=>x=40-16`

`=>x=24`

Vậy `x=24`

__

`3^(x+3) -13=230`

`=> 3^(x+3) = 230+13`

`=>3^(x+3)=243`

`=> 3^(x+3)=3^5`

`=> x+3=5`

`=>x=5-3`

`=>x=2`

Vậy `x=2`

 Câu 20, 21 đk

a. 5x(x-100)=0

=> TH1 : 5x=0 => x= 0

TH2 : x-100 = 0 => x= 100

Vậy .......

b. 34x (2x-6) =0

TH1: 34x=0 => x= 0

TH2; 2x-6=0 => 2x= 6 => x = 3

Vậy ............

a,5x(x-100)=0

=> 5x=0 hoac x-100=0

=> x=0 hoac x=100

b,34x(2x-6)=0

=> 34x=0 hoac 2x-6=0

=>x=0 hoac x=3

c,3x(x+7)-15=27

phan nay nhieu truong hop lam ban dong nao nha goi y

3x=6hoac21

x+7=7hoac 2

va co 4truong hop

phan con lai ko ro ban viet lai di

\(5.\left(x-100\right)=0\)

\(x-100=0\)

\(x=100\)

vay \(x=100\)

\(34.\left(2x-6\right)=0\)

\(2x-6=0\)

\(2x=6\)

\(x=3\)

vay \(x=3\)

\(3.\left(x+7\right)-15=27\)

\(3\left(x+7\right)=42\)

\(x+7=14\)

\(x=7\)

vay \(x=7\)

\(24+3.\left(5-x\right)=0\)

\(24+15-3x=0\)

\(39-3x=0\)

\(3x=39\)

\(x=13\)

vay \(x=13\)

2+4+6+8+10+...+2500

= (2+2500) x 625

= 2502 x 625 = 1563750 =  1250 x 1251

Vậy x=1250

\(x.\left(x+1\right)=2+4+6+...+2500\)

Đặt \(A=2+4+6+...+2500\)

Số số hạng của A:

\(\left(2500-2\right):2+1=1250\) (số)

\(\Rightarrow A=\left(2500+2\right).1250:2=1563750\)

\(\Rightarrow x.\left(x+1\right)=1563750\)

\(x^2+x-1563750=0\)

\(x^2-1250x+1251x-1563750=0\)

\(\left(x^2-1250x\right)+\left(1251x-1563750\right)=0\)

\(x\left(x-1250\right)+1251\left(x-1250\right)=0\)

\(\left(x-1250\right)\left(x+1251\right)=0\)

\(\Rightarrow x-1250=0\) hoặc \(x+1251=0\)

*) \(x-1250=0\)

\(x=1250\)

*) \(x+1251=0\)

\(x=-1251\)

Vậy \(x=-1251;x=1250\)

Chú ý: Nếu đề yêu cầu tìm x là số tự nhiên thì loại x = -1251 và kết luận x = 1250

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