`a)``A=-1/3+1/3^2-1/3^3+1/3^4-...+1/3^50-1/3^51`

`3A=3.(-1/3+1/3^2-1/3^3+1/3^4-...+1/3^50-1/3^51)`

`3A=-1+1/3-1/3^2+1/3^3-...+1/3^49-1/3^50`

`3A+A=(-1+1/3-1/3^2+1/3^3-...+1/3^49-1/3^50)+(-1/3+1/3^2-1/3^3+1/3^4-...+1/3^50-1/3^51)`

`4A=-1-1/3^51`

`A=(-1-1/3^51)/4`

Vậy `A=(-1-1/3^51)/4`

`b)``A=4sqrt(x+2)-1(x>=-2)`

Ta có:`sqrt(x+2)>=0AAx>=-2`

`=>4sqrt(x+2)>=0AAx>=-2`

`=>4sqrt(x+2)-1>=-1AAx>=-2`

Dấu "=" xảy ra `<=>x+2=0`

`=>x=-2`(tmđk)

Vậy `min A=-1` khi `x=-2`

`c)``B=5-1/2sqrt(x-1)(x>=1)`

Ta có:`sqrt(x-1)>=0AAx>=1`

`=>1/2sqrt(x-1)>=0AAx>=1`

`=>-1/2sqrt(x-1)<=0AAx>=1`

`=>5-1/2sqrt(x-1)<=5AAx>=1`

Dấu "=" xảy ra `<=>x-1=0`

`=>x=1`(tmđk)

Vậy `max B=5` khi `x=1`