\(\left(20^2+18^2+16^2+......+4^2+2^2\right)-\left(19^2+17^2+.....+3^2+1^2\right)\)

\(=20^2-19^2+18^2-17^2+......+2^2-1^2\)

\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+.......+\left(2-1\right)\left(2+1\right)\)

\(=39+35+....+7+3\)

\(=\left(39+3\right)\left[\left(39-3\right):4+1\right]:2=210\)

2A = 2^18 - 2^17 - 2^15 - 2^14 - 2^13 - ... - 2^2 - 2

2A - A = A = ( 2^18 - 2^17 - 2^16 - 2^15 - 2^14 - 2^13 - ... - 2^2 - 2 ) - ( 2^17 - 2^16 - 2^15 - 2^14 - 2^13 - ... -2 - 1 )

A = 2^18 - 2^17 - 2^16 - 2^15 - 2^14 - 2^13 - ... - 2^2 - 2 - 2^17 + 2^16 + 2^15 +  2^14 + 2^13 + ... + 2 + 1

A = 2^18 - 2^17 - 2^17 + 1 

A = 2^18 - 2 . 2^17 + 1 

A =2^17- 2^16-...-2-2^0

Ax2= 2^18-2^17-.....-2^2-2^1

Ax2-A=2^18-2^17-...-2^2-2^1-2^17+2^16+....+2^1+1

A=2^18-2^17-2^17+1

A=2^18-2^17X2+1

A=2^18-2^18+1

A=0+1

A=1

Bài 1: 

Ta có: 

\(A=9x^4-15x^3-6x^2+5=3x^2\left(3x^2-5x\right)-6x^2+5=3x^2.2-6x^2+5=6x^2-6x^2+5=5\)

Vậy,  \(A=5\)

Bài 2: Ta có:

\(3^{15}+3^{16}+3^{17}=3^{15}+3^{15}.3+3^{15}.3^2=3^{15}.\left(1+3+3^2\right)=3^{15}.13\)

\(\Rightarrow3^{15}.13\)  chia hết cho  \(13\)

Do đó:  \(3^{15}+3^{16}+3^{17}\)  chia hết cho  \(13\)

2, 

a) \(315-\left(135-x\right)=215\)

\(\Rightarrow135-x=315-215\)

\(\Rightarrow135-x=100\)

\(\Rightarrow x=135-100\)

\(\Rightarrow x=35\)

b) \(x-320:32=25\cdot16\)

\(\Rightarrow x-10=5^2\cdot4^2\)

\(\Rightarrow x-10=20^2\)

\(\Rightarrow x-10=400\)

\(\Rightarrow x=410\)

c) \(3\cdot x-2018:2=23\)

\(=3\cdot x-1009=23\)

\(\Rightarrow3\cdot x=1032\)

\(\Rightarrow x=1032:3\)

\(\Rightarrow x=344\)

d) \(280-9\cdot x-x=80\)

\(\Rightarrow280-x\cdot\left(9+1\right)=80\)

\(\Rightarrow280-10\cdot x=80\)

\(\Rightarrow10\cdot x=280-80\)

\(\Rightarrow10\cdot x=200\)

\(\Rightarrow x=20\)

e) \(38\cdot x-12\cdot x-x\cdot16=40\)

\(\Rightarrow x\cdot\left(38-12-16\right)=40\)

\(\Rightarrow x\cdot10=40\)

\(\Rightarrow x=40:10\)

\(\Rightarrow x=4\)

bài 1 cậu ko giải giúp mình dc à

Ta có : ( 1 X 3 X .........X6) chia hết cho 2 ( số chẵn)

( 18 x 17 x16 x 15 ) chia hết cho 2 ( số chẴN ) 

SUY RA  biểu thức trên = chẵn - chẵn = chẵn ( chia hết cho 2 ) 

ta có: 9 chia hết cho 2;3;9

=> 1 x 3 x 5 x 7 x 9 x ...x 51 x 53 x 6 chia hết cho 2;3;9

18 chia hết cho 2;3;9

=> 18 x 17 x 16 x 15 chia hết cho 2;3;9

=> ( 1 x 3 x 5 x 7 x ...x 51 x 53 x 6) - ( 18 x 17 x 16 x 15) chia hết cho 2;3;9

ta có: 5 chia hết cho 5

=> 1 x 3 x 5 x 7 x ....x 51 x 53 x 6 chia hết cho 5

15 chia hết cho 5

=> 18 x 17 x 16 x 15 chia hết cho 5

=> ( 1 x 3 x5 x 7 x ...x 51 x 53 x 6) - ( 18 x17 x 16 x15) chia hết cho 5

KL: ( 1 x 3 x 5 x7 x...x51 x 53 x 6) - ( 18 x17 x 16 x 15) chia hết cho 2;3;5;9

`B17:`

`a)` Với `x \ne +-3` có:

`A=[x+15]/[x^2-9]+2/[x+3]`

`A=[x+15+2(x-3)]/[(x-3)(x+3)]`

`A=[x+15+2x-6]/[(x-3)(x+3)]`

`A=[3x+9]/[(x-3)(x+3)]=3/[x-3]`

`b)A=[-1]/2<=>3/[x-3]=-1/2<=>-x+3=6<=>x=-3` (ko t/m)

   `=>` Ko có gtr nào của `x` t/m

`c)A in ZZ<=>3/[x-3] in ZZ`

   `=>x-3 in Ư_3`

 Mà `Ư_3={+-1;+-3}`

`@x-3=1=>x=4`

`@x-3=-1=>x=2`

`@x-3=3=>x=6`

`@x-3=-3=>x=0`

________________________________

`B18:`

`a)M=1/3`             `ĐK: x  \ne +-4`

`<=>(4/[x-4]-4/[x+4]).[x^2+8x+16]/32=1/3`

`<=>[4(x+4)-4(x-4)]/[(x-4)(x+4)].[(x+4)^2]/32=1/3`

`<=>32/[x-4].[x+4]/32=1/3`

`<=>3x+12=x-4`

`<=>x=-8` (t/m)

Bài 4:

\(A=2x^2-15\ge-15\\ A_{min}=-15\Leftrightarrow x=0\\ B=2\left(x+1\right)^2-17\ge-17\\ B_{min}=-17\Leftrightarrow x=-1\)

Bài 5:

\(A=-x^2+14\le14\\ A_{max}=14\Leftrightarrow x=0\\ B=25-\left(x-2\right)^2\le25\\ B_{max}=25\Leftrightarrow x=2\)