Answer:

Câu `5` :

`a) (x+2)/(-2)=5/10`

`=> (x+2)/(-2)=1/2`

`=> (x+2)*2= 1*(-2)`

`=> 2x+4=-2`

`=> 2x=-2-4`

`=> 2x=-6`

`=>x= (-6):2`

`=> x= -3`

Vậy `x=-3`

`b) 1/9 + (x-1/3)^2=5/9`

`=> (x-1/3)^2= 5/9-1/9`

`=> (x-1/3)^2= 4/9`

`=> (x-1/3)^2= (±2/3)^2`

`=> x-1/3= 2/3` hoặc `x-1/3=-2/3`

`=> x= 2/3 + 1/3` hoặc `x=-2/3+1/3`

`=> x= 3/3` hoặc `x= -1/3`

`=>x=1` hoặc `x=-1/3`

Vậy `x∈{1;-1/3}`