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\(\left(20^2+18^2+16^2+......+4^2+2^2\right)-\left(19^2+17^2+.....+3^2+1^2\right)\)
\(=20^2-19^2+18^2-17^2+......+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+.......+\left(2-1\right)\left(2+1\right)\)
\(=39+35+....+7+3\)
\(=\left(39+3\right)\left[\left(39-3\right):4+1\right]:2=210\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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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 \(13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`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)
![](https://rs.olm.vn/images/avt/0.png?1311)
x=11 suy ra 12=x+1 thay vào A ta có:
A=x^17- (x+1)x^16 + (x+1)x^15 - (x+1)x^14 + .....- (x+1)x^2+(x+1)x -1
= x^17 - x^17 -x^16 + x^16 + x^15 - x^15 - x^14 +.....- x^3 -x^2 + x^2 +x -1
= x-1= 11-1=10
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=2^{17}-\left(1+2+2^2+...+2^{16}\right)\)
đặt \(1+2+2^2+...+2^{16}=B\Rightarrow A=2^{17}-B\)
\(B=1+2+2^2+...+2^{16}\)
\(2B=2+2^2+2^3+...+2^{17}\)
\(B=2B-B=\left(2+2^2+...+2^{17}\right)-\left(1+2+...+2^{16}\right)\)
\(B=2^{17}-1\)
\(A=2^{17}-B=2^{17}-\left(2^{17}-1\right)=2^{17}-2^{17}+1=1\)
Vậy A=1
A=214 + 1