Đáp án + Giải thích các bước giải:

`e)`

`(x+y)^2-(x-y)^2+(x+y)(x-y)`

`=(x^2+2xy+y^2)-(x^2-2xy+y^2)+x^2-y^2`

`=x^2+2xy+y^2-x^2+2xy-y^2+x^2-y^2`

`=(x^2-x^2+x^2)+(y^2-y^2-y^2)+(2xy+2xy)`

`=x^2-y^2+4xy`

`f)`

`12(2x-5)^2-3(1+4x)(4x-1)`

`=12(4x^2-10x+25)-3(16x^2-1)`

`=48x^2-120x+300-48x^2+3`

`=(48x^2-48x^2)-120x+(300+3)`

`=303-120x`

`g)`

`(a+3)(a-3)(a^2+3a+9)(a^2-3a+9)`

`=[(a+3)(a^2-3a+9)][(a-3)(a^2+3a+9)]`

`=(a^3+3^3)(a^3-b^3)`

`=(a^3+27)(a^3-27)`

`=a^6-27a^3+27a^3-729`

`=a^6-729`

`h)`

`100^2-99^2+98^2-97^2+...+2^2-1^2`

`=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)`

`=100+99+98+97+...+2+1`

`=(100+1).100:2`

`=5050`

`i)`

`3(2^2+1)(2^4+1)(2^8+1)...(2^64+1)+1`

`=(2^2-1)(2^2+1)(2^4+1)(2^8+1)...(2^64+1)+1`

`=(2^4-1)(2^4+1)(2^8+1)...(2^64+1)+1`

`=(2^8-1)(2^8+1)...(2^64+1)+1`

`=(2^{16}-1)(2^{16}+1)...(2^64+1)+1`

`=(2^{32}-1)(2^{32}+1)...(2^64+1)+1`

`=(2^{64}-1)(2^64+1)+1`

`=2^{128}-1+1`

`=2^{128}`

$\color{darkgoldenrod}{e)}\ (x+y)^2-(x-y)^2+(x+y)(x-y)$

$=(x+y-x+y)(x+y+x-y)+(x^2-y^2)$

$=2x.2y+x^2-y^2$

$=4xy+x^2-y^2$

$=4xy+x^2-y^2$

$\\$$\\$

$\color{darkgoldenrod}{f)}\ 12(2x-5)^2-3(1+4x)(4x-1)$

$=12(4x^2-20x+25)-3(16x^2-1)$

$=48x^2-240x+300-48x^2+3$

$=303-240x$

$\\$$\\$

$\color{darkgoldenrod}{g)}\ (a+3)(a-3)(a^2+3a+9)(a^2-3a+9)$

$=[(a+3)(a^2-3a+9)][(a-3)(a^2+3a+9)]$

$=(a^3+27)(a^3-27)$

$=(a^3)^2-27^2$

$=a^6-729$

$\\$$\\$

$\color{darkgoldenrod}{h)}\ 100^2-99^2+98^2-97^2+...+2^2-1^2$

$=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)$

$=(100+99)+(98+97)+...+(2+1)$

$=100+99+98+97+...+2+1$

$=(100+1).100:2$

$=101.50$

$=5050$

$\\$$\\$

$\color{darkgoldenrod}{i)}\ 3(2^2+1)(2^4+1)(2^8+1)...(2^{64}+1)+1$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)...(2^{64}+1)+1$

$=(2^4-1)(2^4+1)(2^8+1)...(2^{64}+1)+1$

$=(2^8-1)(2^8+1)...(2^{64}+1)+1$

$=(2^{16}-1)...(2^{64}+1)+1$

$=(2^{64}-1)(2^{64}+1)+1$

$=(2^{128}-1)+1$

$=2^{128}-1+1$

$=2^{128}$