Đáp án:
Giải thích các bước giải:
`x^2+y^2=6(x-y-3)`
`=>x^2+y^2-6(x-y-3)=0`
`=>(x^2-6x+9)+(y^2+6y+9)=0`
`=>(x-3)^2+(y+3)^2=0`
Ta có
`(x-3)^2+(y+3)^2>=0 ∀x,y`
Mà `(x-3)^2+(y+3)^2=0`
`=>(x-3)^2=(y+3)^2=0`
`=>x=3,y=-3`
`=>B=x^2019+y^2019+(x+y)^2020+2021`
`=3^2019+(-3)^2019+[3+(-3)]^2020+2021`
`=0+0+2021`
`=2021`
`=>B=2021`
`b)`
`A=(5^2+1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)(5^64+1)`
`=>24A=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)(5^64+1)`
`=>24A=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)(5^64+1)`
`=>24A=(5^4-1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)(5^64+1)`
`=>24A=(5^8-1)(5^8+1)(5^16+1)(5^32+1)(5^64+1)`
`=>24A=(5^16-1)(5^16+1)(5^32+1)(5^64+1)`
`=>24A=(5^32-1)(5^32+1)(5^64+1)`
`=>24A=(5^64-1)(5^64+1)`
`=>24A=5^128-1`
`=>A=(5^128-1)/24`
